In this article, we shall study change in enthalpy for different chemical processes.

Enthalpy of Formation (Δ_fH° or Δ_formationH°):

The change in enthalpy of a chemical reaction at a given temperature and pressure, when one mole of the substance is formed from its constituent elements in their standard states is called the heat of formation.

Explanation: Consider the following reaction

C_(s) + O_2(g)  → CO_2(g)   ,  Δ_formationH°  =  -395.39 kJ mol^-1

Since one mole of carbon dioxide gas is formed we can say that the heat of formation of carbon dioxide gas is -395.39 kJ.

Notes:

• The standard state of the element is that stable state of the element in which it exists at 1 atm. Pressure and 298 K
• Enthalpies of elements in their standard states are arbitrarily taken as zero.
• Enthalpy of a compound is equal to its heat of formation.
• When solving problems on the heat of formation make sure that the product side of the thermochemical equation has one mole of the substance whose heat of formation is to be calculated.

Enthalpy of Dissociation (Δ_dissociationH°):

Change in enthalpy of a chemical reaction at a given temperature and pressure, when one mole of a substance is dissociated into its constituent elements is called the heat of dissociation.

Explanation: Consider the following reaction

H_2(g) →  2H_(g),    Δ_dissociationH° = + 435.136 kJ mol^-1

Since one mole of hydrogen gas is dissociated the heat of dissociation of hydrogen gas is + 435.136 kJ

Enthalpy of Combustion (Δ_cH° or Δ_combustionH°):

Change in the enthalpy of a chemical reaction at a given temperature and pressure when one mole of a substance is combusted (burn) completely in excess of oxygen is called the heat of combustion.

Explanation: Consider the following reaction

C_(s) + O_2(g)  → CO_2(g)   , ΔH  =  -395.39 kJ mol^-1

Since one mole of carbon is combusted completely the heat of combustion of carbon is – 395.39 kJ.

Enthalpy of Neutralization (Δ_{neutralization}H°):

Change in the enthalpy of a chemical reaction at a given temperature and pressure when one gram equivalent weight of acid is completely neutralized by one gram equivalent weight of the base is called the heat of neutralization.

Explanation: Consider following reaction

HCl_(aq) + NaOH_(aq)  →   NaCl  +    H₂O     Δ_{neutralization}H°  =  -56.9 kJ

The heat of neutralization of HCI by NaOH is -56.9 KJ.

For all strong acids and bases, the heat of neutralization is the same because, in their neutralization reaction, there is a combination of H+ ions of an acid with OH- ions of the base to produce un-dissociated water.

Change of Phase:

Change of phase involves the change in the physical state of matter. During the phase change, the chemical properties of the substance do not change but physical properties change. The following are the types of phase changes.

Fusion: This is the process in which the matter changes from solid-state to liquid state. It is endothermic process.

e.g. melting of ice H₂O_(s) → H₂O_(l)

Vapourization: This is the process in which the matter changes from liquid state to gaseous state. It is an endothermic process.

e.g. boiling of water H₂O_(l) → H₂O_(g)

Sublimation: This is the process in which the matter changes from the solid-state into a gaseous state directly. It is an endothermic process.

e.g. heating of camphor Camphor_(s) → Camphor_(g)

Characteristics of Change of Phase:

• The phase change always takes place at constant pressure and temperature.
• During the phase transition, there is an equilibrium between the two phases. Thus both the phases exist simultaneously.
• The Change in temperature takes place only when completion of phase transition.

Enthalpy of Fusion (Δ_fusH°):

The enthalpy-change that accompanies the fusion of one mole of a solid without the change in temperature at constant pressure is called its enthalpy of fusion.

Explanation: Consider the following reaction

H₂O_(s) → H₂O_(l),               Δ_fusionH  = +6.01  kJ mol^-1

Thus, the equation indicates that when one mole of ice melts at 0° C at 1 atm pressure, the enthalpy-change is 6.01 kJ. i.e. 6.01 kJ of energy is absorbed.

Enthalpy of Freezing (Δ_freezeH°):

The enthalpy change that accompanies the freezing of one mole of a liquid without a change in temperature at constant pressure is called its enthalpy of freezing.

Explanation: Consider the following reaction

H₂O_(l) → H₂O_(s),               Δ_freezeH  = +6.01  kJ mol^-1

Thus, the equation indicates that when one mole of water freezes at 0° C at 1 atm pressure, the change in enthalpy is -6.01 kJ. i.e. 6.01 kJ of energy is released.

Enthalpy of Vaporization (Δ_vaporizationH°):

The enthalpy change that accompanies the vaporization of one mole of a liquid without a change in temperature at constant pressure is called its enthalpy of vaporization.

Explanation: Consider the following reaction

H₂O_(l) → H₂O_(g),   Δ_{vapourization}H  = + 40.7  kJ mol^-1

Thus, the equation indicates that when one mole of water vaporizes at 100° C at 1 atm pressure, the change in enthalpy is + 40.7 kJ. i.e. 40.7 kJ of energy is absorbed.

Enthalpy of Condensation (Δ_condensationH°):

The enthalpy change that accompanies the condensation of one mole of a liquid without a change in temperature at constant pressure is called its enthalpy of condensation.

Explanation: Consider the following reaction

H₂O_(l) → H₂O_(g),   Δ_condensationH  = – 40.7  kJ mol^-1

Thus, the equation indicates that when one mole of water vapours condense at 100° C at 1 atm pressure, the enthalpy-change is – 40.7 kJ. i.e. 40.7 kJ of energy is released.

Enthalpy of Sublimation (Δ_sublimationH°):

The direct conversion of solid to vapour without going through the liquid state is called sublimation. The enthalpy-change that accompanies the condensation of one mole of a solid directly into vapours at a constant temperature at constant pressure is called its enthalpy of sublimation.

Explanation: Consider the following reaction

H₂O_(s) → H₂O_(g),       Δ_sublimationH  = +51.08  kJ mol^-1

Thus, the equation indicates that when one mole of ice sublimes at O° C at 1 atm pressure, the enthalpy change is + 51.08 kJ. i.e. 51.8 kJ of energy is absorbed.

It is to be noted that Δ_sublimationH = Δ_fusionH + Δ_{vapourization}H

Atomic or Molecular Changes:

Enthalpy of Ionization (Δ_ionizationH°):

The enthalpy change that accompanies the removal of an electron from each atom or ion in one mole of gaseous atoms or ions is called its enthalpy of ionization.

Explanation: Consider the following reaction

Na_(g)   →   Na⁺_(g)+ e^– ,    Δ_ionizationH  = 494  kJ  mol^-1

Thus, the equation indicates that when one mole of gaseous sodium atom ionizes to Na⁺_(g) ion, the change in enthalpy is + 494 kJ. i.e. 494 kJ of energy is absorbed.

Enthalpy of Atomization (Δ_atomizationH°):

The enthalpy change that accompanies the dissociation of all the molecules in one mole of gas-phase substance into gaseous atoms is called its enthalpy of atomization.

Explanation: Consider the following reaction

Cl_2(g) → Cl_(g)+ Cl_(g),          Δ_atomizationH  = 242  kJ mol^-1

Thus, the equation indicates that when one mole of gaseous chlorine molecule dissociates completely into its atomic form in the gaseous state then the change in enthalpy is + 242 kJ. i.e. 242 kJ of energy is absorbed.

Enthalpy of Solution (Δ_solutionH°):

Change in enthalpy of chemical reaction at a given temperature and pressure, when one mole of a solution is dissolved in a specified quantity of solvent so as to form a solution of particular concentration is called as enthalpy of a solution.

Explanation: Consider the following reaction

KOH_(s)   +   H₂O_(l) → KOH_(aq)        Δ_solutionH   = -58.57 KJ mol^-1

Thus, the equation indicates that when one mole of potassium hydroxide (solute) dissolves in one mole of water (solvent) to form one mole of potassium hydroxide solution in water then the change in enthalpy is -58.57 kJ. i.e. 58.57 kJ of energy is released

Bond Enthalpy (Bond Energy):

Chemical reactions involve the breaking and making of chemical bonds. Energy is required to break a bond and energy is released when a bond is formed. It is possible to relate the heat of reaction to changes in energy associated with the breaking and making of chemical bonds. With reference to the enthalpy changes associated with chemical bonds, two different terms are used in thermodynamics. (i) Bond dissociation enthalpy (ii) Mean bond enthalpy. Let us discuss these terms with reference to diatomic and polyatomic molecules.

Diatomic Molecules:

Consider the following process in which the bonds in one mole of dihydrogen gas (H2) are broken:

H_2(g) →  2H_(g) ;   ΔH–HHO = 435.0 kJ mol^-1

The enthalpy change involved in this process is the bond dissociation enthalpy of H–H bond.

The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase. Note that it is the same as the enthalpy of atomization of dihydrogen. This is true for all diatomic molecules.

Polyatomic Molecule:

In the case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule. Let us consider a polyatomic molecule like methane, CH4.  The overall thermochemical equation for its atomization reaction is given below:

CH_4(g) →  C_(g) + 4H_(g) , ΔH = +1665 KJ.

In methane, all the four C – H bonds are identical in bond length and energy. However, the energies required to break the individual C – H bonds in each successive step differ. In such cases, we use mean bond enthalpy

CH_4(g) →  CH_3(g) + H_(g) , ΔH = +427 KJ.

CH_3(g) → CH_2(g) + H_(g), ΔH = +439 KJ

CH_2(g) → CH_(g) + H_(g), ΔH = +452 KJ

CH_(g) → C (g) + H_(g),   ΔH = +347 KJ

Adding above reactions we get

CH_4(g) →  C_(g) + 4H_(g) , ΔH = +1665 KJ

We find that mean C–H bond enthalpy in methane as 1664/4 = 416 kJ/mol. Using Hess’s law, bond enthalpies can be calculated.

Enthalpy of Reaction from Bond Enthalpy:

The reaction enthalpies are very important quantities as these arise from the changes that accompany the breaking of old bonds and the formation of the new bonds. We can predict the enthalpy of a reaction in the gas phase if we know different bond enthalpies. The standard enthalpy of reaction is related to bond enthalpies of the reactants and products in gas phase reactions as:

ΔH = ∑ Bond enthalpies _reactants  –     ∑ Bond enthalpies _products

Remember that this relationship is approximate and is valid when all substances (reactants and products) in the reaction are in a gaseous state.

The values of given bond enthalpy can be used to calculate bond enthalpies of specific bonds in the molecule.

Crystal Lattice Energy (Δ_LatticeH°):

Crystal lattice energy is defined as the enthalpy change or energy released accompanying formation of 1 mole of crystalline solid from its constituent ions in the gaseous state at a constant temperature.

Explanation:

M⁺_(g)  + X^–_(g) → M⁺X^–_{(s) ,}  ΔH = – x KJ  mol^-1

Thus, the equation indicates that when one mole of ionic compound M+X-(s) is formed from its constituent ions the change in enthalpy is – x kJ i.e. x kJ of energy is evolved. Crystal lattice energy is always negative.

The sequence of actions involved in the formation of 1 mole of an ionic compound in its standard state from its constituent elements in their states at constant temperature and pressure is called Born-Haber cycle.

Factors Affecting Crystal Lattice Energy:

Crystal lattice energy depends on the interionic distance in the crystalline solid. As the distance decreases, the crystal lattice energy increases. Crystal lattice energy depends on the charge of constituent cations and anions.

Enthalpy of Hydration (Δ_HydrationH°):

It is defined as the heat evolved when one mole of gaseous ions dissolve in water by hydration to give infinitely dilute solution at constant temperature and pressure.

Explanation: Consider the following reaction

Na⁺_(g) + aq → Na⁺_(ag) ,     ΔH = – 390 KJ mol^-1

Thus, the equation indicates that when one mole of sodium ion in the gaseous state is dissolved in water the change in enthalpy is – 390 kJ. i.e. 390 kJ of energy is released.

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The branch of chemistry which deals with the quantitative study of thermal or heat changes in various chemical reactions is known as thermochemistry. In this article, We shall discuss very important concept of chemistry i.e. heat of reaction.

Thermochemical Equation:

An equation which indicates the heat changes in a chemical reaction at a certain temperature and pressure with an indication of states of reactants and products is called a thermochemical equation

Example:

C_(s)  + O_2(g)  → CO_2(g)   , ΔH  =  -395.39 kJ

This thermochemical equation indicates that when one mole of solid carbon reacts with one mole of gaseous oxygen at constant pressure one mole of gaseous carbon dioxide is obtained. During this reaction, 395.39 kJ of heat is evolved.

Guidelines for Writing Thermochemical Equation:

• The equation must be automatically balanced like a chemical equation.
• The value of ΔH must be indicated. ΔH is negative for exothermic reaction and positive for an endothermic reaction.
• The physical state of each reactant and each product must be indicated. Symbols such as (s) for solid-state, (l) for the liquid state, (g) for gaseous state and (aq.) for an aqueous solution should be indicated.
• The thermochemical equation can be reversed. During the reversal of the sign of ΔH must be changed.
• The temperature of the reaction can be written as a suffix to ΔH

Example:

C_(s)  + O_2(g)  → CO_2(g)   , ΔH  =  -395.39 kJ

This thermochemical equation indicates that when one mole of solid carbon reacts with one mole of gaseous oxygen at constant pressure one mole of gaseous carbon dioxide is obtained. During this reaction, 395.39 kJ of heat is evolved.

The necessity of Mentioning of State of a Substance:

It is important to mention the physical state of substances in the thermochemical equation because the change of physical state is also accompanied by the enthalpy change.

Example: Consider the following thermochemical equations

H_2(g) +    1/2O_2(g)  →    H₂O_(l) , ΔH =  – 286 kJ

Thus, when 1 mole of hydrogen gas reacts with half a mole of oxygen gas to form one mole of liquid water, 286 kJ of heat is produced.

H_2(g) +    1/2O_2(g)  →    H₂O_(g) , ΔH =  – 249 kJ

Thus, when 1 mole of hydrogen gas reacts with half a mole of oxygen gas to form one mole of water vapours, 249 kJ of heat is produced. Hence, whenever there is a state change, there is a change in enthalpy of reaction. Hence in thermochemical reaction, the state of each substance involved should be mentioned.

Heat of Reaction OR Enthalpy of Chemical Reaction:

The difference between the sum of enthalpies of products and the sum of enthalpies of reactants at a given temperature is called as the heat of reaction. It is denoted by ΔH or ΔU

Explanation:

Consider a general chemical reaction 

A    +    B →  C     +   D.

Let H_A , H_B, H_C, and H_D be the enthalpies of A, B, C and D respectively then the heat of reaction is given by

ΔH  =   (H_C +  H_D)  –  (H_A +  H_B)

The heat of reaction can be determined either at constant pressure or at constant volume.

Heat of Reaction at Constant Pressure:

The difference between the sum of enthalpies of products and the sum of enthalpies of reactants at a given temperature and constant pressure is called the heat of reaction at the constant pressure at a given temperature.

It is denoted by ΔH. Normally heat of reaction at constant pressure is specified at 298 K and 1 atm. Pressure. This heat of reaction is called the standard heat of reaction.

Thus at constant pressure heat of reaction is given by

ΔH = ∑ ΔH_Products  –     ∑ ΔH_Reactants

ΔH is negative for exothermic reaction and positive for an endothermic reaction.

Heat of Reaction at Constant Volume:

The difference between the sum of internal energies of products and the sum of internal energies of reactants at a given temperature and constant volume is called the heat of reaction at the constant volume at a given temperature. It is denoted by ΔE.

Thus at constant pressure heat of reaction is given by

ΔU = ∑ ΔU_Products  –     ∑ ΔU_Reactants

Factors Affecting Heat of Reaction:

• Physical states of substances involved.
• The amount of substance involved.
• Way of carrying out the reaction i.e. in a case of gaseous reactions heat of reaction depends on whether the reaction is carried out at constant pressure or at constant volume.
• The pressure of reactants and products.
• The temperature (as explained by Kirchhoff’s equations)

Different Types of Chemical Reactions on the Basis of Change in the Enthalpy:

On the basis of change in the enthalpy chemical reactions are classified into exothermic reactions and endothermic reactions

Exothermic Reactions:

The chemical reactions which involve the evolution of heat are called as exothermic reactions.

Example:

C_(s)+ O_2(g)    →  CO_2(g) , ΔH = -395.39 kJ

In this case, the enthalpy of products is less than the enthalpy of reactants. For such reactions, the change in enthalpy is always negative.

Characteristics of Exothermic Reactions:

• The chemical reactions which involve the evolution of heat are called exothermic reactions.
• For such reactions, the change in enthalpy is always negative.
• In this case, the enthalpy of products is less than the enthalpy of reactants
• Products are more stable than the reactants.

Endothermic Reactions:

The chemical reactions which involve absorption of heat are called endothermic reactions.

Example :

2C_(s)+ H_2(g)    →  C₂H_2(g) , ΔH = + 225.94kJ

In this case, the enthalpy of products is more than the enthalpy of reactants. For such reactions, the change in enthalpy is always positive.

Characteristics of Endothermic Reactions:

• The chemical reactions which involve absorption of heat are called endothermic reactions.
• For such reactions, the change in enthalpy is always positive.
• In this case, the enthalpy of products is more than the enthalpy of reactants.
• Products are less stable than the reactants.

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Science > Chemistry > Chemical Thermodynamics and Energetics > Heat of Reaction

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The enthalpy of system is defined as the sum of the internal energy of the system and energy that arises due to pressure and volume. It is denoted by letter H. Mathematically,

H = U + PV

Where, H = Enthalpy U = Internal Energy P = Pressure V = Volume

As the internal energy E, Pressure P and Volume V are state functions, the enthalpy of the system is a state function.

Change in Enthalpy of System:

Let H₁ and H₂ be the enthalpies of the system in the initial state and final state respectively. Let P₁, V_1, and U1 be the pressure, volume and internal energy of the system in the initial state. Let P2, V2, and U₂ be the pressure, volume and internal energy of the system in the final state.

The change in enthalpy is given by ΔH = H₁ – H₂         ……………….    (1)      

By definition of enthalpy   H = U + PV

For initial state,       H1 = U₁ + P ₁V₁

For final state,     H₂ = U₂ + P₂ V₂

Substituting these values in equation (1)

Δ H =   (U₂ + P₂ V₂) – (U₁ + P₁V₁)   

∴ Δ H =   (U₂ – U₁) + (P₂ V₂ – P₁ V₁)

At constant pressure P₁    =   P₂ =   P 

∴ Δ H =   (U2 – U₁) + (P V₂ – P V₁) 

∴ Δ H =    (U₂ – U₁) + P (V₂ – V₁)

∴ Δ H   = ΔU + PΔ V

This is mathematical expression for change in enthalpy of system.

Thus the change in enthalpy of the system is equal to the sum of the increase in the internal energy of the system and the mechanical work due to expansion.

If Δ H is positive then the reaction is endothermic and when ΔH is negative the reaction is exothermic.

Change in Enthalpy of System at Constant Pressure (Isobaric Process):

The expression for the change in enthalpy of a system at constant pressure is

Δ H = Δ U   + PΔ V     …………  (1)

Where, ΔH = Change in enthalpy ΔU = Change in internal energy P = Pressure ΔV = Change in volume

The mathematical statement of the first law of thermodynamics is  

ΔU = q + PΔ V 

Where,   q = Heat supplied to the system ΔU = Change in internal energy P = Pressure ΔV = Change in volume

Let us denote q = q_p = heat absorbed by the system at constant pressure.

q_p = ΔU + PΔV    ……….   (2)  

From equations (1) and (2) we have

ΔH = q_p

Thus at constant pressure, the change in enthalpy is equal to heat absorbed at constant pressure.

Change in enthalpy of System at Constant Volume (Isochoric Process):

The expression for the change in enthalpy of a system is

Δ H = Δ U + PΔ V

Where, ΔH = Change in enthalpy ΔU = Change in internal energy P = Pressure ΔV = Change in volume

In isochoric process ΔV = 0

∴   Δ H = Δ U +  P(0)

Δ H = Δ U    ………….. (1)

The mathematical statement of the first law of thermodynamics is

Δ U = q + PΔV ………….. (2)

Where,   q = Heat supplied to the system ΔU = Change in internal energy P = Pressure ΔV = Change in volume

Substituting ΔV = 0 in equation (2)

q = ΔU

let us denote q = q_v = heat absorbed by system at constant volume.

q_v = ΔE   …………..  (3)

From equations (1) and (3) we have

ΔH = ΔU = q_v

Thus at constant volume, the change in enthalpy is equal to the heat absorbed at constant volume and also equal to change in internal energy. Thus in the isochoric process heat supplied to the system is used for increasing the internal energy of the system.

The Relation between ΔH and ΔU:

Change in the enthalpy of a system is given by

Δ H = ΔU   + PΔV ……(1)

Where, ΔU = Change in internal energy P = Pressure of the system ΔV = Change in volume

But for reactions involving gases

PV = nRT

Where P = Pressure of a gas V = Volume of the gas n = Number of moles of the gas

R = Universal gas constant T = Absolute temperature of the gas

For initial state,    P V₁     = n₁RT

For Final state,    P V₂       = n₂RT

PV₂ – PV₁   =    n₂RT – n₁RT

∴ P( V₂ –  V₁)  =   (n₂ –  n₁ )RT

∴   PΔV = ΔnRT ….. (2)

Substituting in equation (1)

Δ H = ΔU   +  ΔnRT

WhereΔH =   Change in enthalpy or heat of reaction at constant pressure.

ΔU =   Change in internal energy or heat

Δn = Difference in the number of moles of gaseous products and reactants.

R   = Universal gas constant 8.314 J/ mol / k,      T   =   Absolute temperature.

Work Done in a Chemical Reaction:

The work done by a system at constant pressure and temperature is given by

W =   – P_extΔV

Assuming P_ext = P

W = – P( V₂  –  V₁)

∴ W =   – PV₂ + PV₁

But for reactions involving gases    PV = nRT

Where, P = Pressure of a gas V = Volume of the gas n = Number of moles of the gas

R = Universal gas constant T = Absolute temperature of the gas

For initial state,    P V₁ = n₁RT

For Final state,    P V₂       = n₂RT

W = – PV₂ + PV₁ =    – n₂RT + n₁RT

∴ W =   – (n₂ –  n₁ )RT

∴   W = – ΔnRT

This is an expression for work done in chemical reaction.

Conditions Under Which Change in Enthalpy of System (ΔH) is Equal to Change in Internal Energy of the System (ΔU):

When the reaction is carried out in a closed vessel, there is no change in volume. ΔV = 0, hence Δ H = ΔU + PΔV gives Δ H = ΔU.

When the reaction involves only solids and liquids, then change in volume is negligible. ΔV = 0, hence Δ H = ΔU + PΔV gives Δ H = ΔU.

In a chemical reaction in which the number of gaseous reactants consumed is equal to the number of moles of gaseous product formed then

Δn = 0, hence ΔH =ΔU + ΔnRT gives ΔH = ΔU.

Notes: 

• Work has significance only when the number of gaseous reactants consumed and the number of moles of gaseous products formed is different and there is a change in volume.
• Heat Supplied q is not thermodynamic function but heat supplied at constant pressure (qp) and heat supplied at constant volume (qv) are thermodynamic functions:
• By the first law of thermodynamics, ΔU = q + W.  In this relation ΔU is a state function as well as thermodynamic function. Work is path function as well as non-thermodynamic function. For algebraic addition q on the right-hand side of the equation, heat supplied is path function as well as non-thermodynamic function.
• For a constant pressure process, ΔH = qp Where, ΔH is the enthalpy of a system which is state function as well as thermodynamic function. Hence heat supplied at constant pressure qp is a thermodynamic function.
• For a constant volume process, ΔU = q_p Where, ΔU is the internal energy of a system which is state function as well as thermodynamic function. Hence heat supplied at constant volume q_v is a thermodynamic function.

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Science > Chemistry > Chemical Thermodynamics and Energetics > Enthalpy of a System

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In this article, we shall study the first law of thermodynamics and its application to different chemical processes.

Zeroth Law of Thermodynamics:

Statement:

If two bodies (say A and B) are in thermal equilibrium of the third body (say C) then body A and B will also be in thermal equilibrium with each other

Significance of the Law:

The common use of the thermometer in comparing the temperature of any two or more systems is based on the zeroth law of thermodynamics.

First Law of Thermodynamics:

Statement:

Different forms of the first law of thermodynamics are as follows

• Energy can’t be created nor destroyed but it can be converted from one form into the other (or forms) or into work.
• When a quantity of energy of one kind disappears, then an equivalent amount of energy of another kind makes its appearance.
• It is impossible to make a perpetual motion machine which would produce work without consuming energy.
• The total amount of energy of an isolated system remains constant, it may change from one form to another.
• The energy of the universe remains constant.
• For a system in contact with the surroundings, the sum of energies of the system and its surroundings remains constant however differently it may be shared between the two.

Mathematical Statement of the First Law:

Consider the system in state I (initial state) with internal energy U₁. It is converted into state II (final state) with internal energy U₂ by supplying ‘q’ amount of heat to it. During this process, some work “W” is done by the system on the surroundings. Heat absorbed by the system is used for a) increasing internal energy of a system and b) to do some mechanical work “W”. 

Now, Change in the internal energy of the system is equal to the heat supplied plus Work done.

Thus, Final internal energy = U₂ = U₁ +   q +

∴   (U₂ – U₁)   =      q + W

∴   ΔU = q   + W

In the pressure-volume type of work W = PΔV

∴   ΔU = q +  PΔV

This is the mathematical equation of the first law of thermodynamics.

Sign Conventions for q, W, and ΔU:

• When heat is absorbed by the system q is positive.
• When heat is rejected or given out by the system q is negative.
• When the work is done on the system by surroundings (Work of compression) then W is positive.
• When the work is done by the system on the surroundings (Work of expansion) then, W is negative.
• When there is an increase in the internal energy of the system (increase in temperature) ΔU is positive.
• When there is a decrease in the internal energy of the system (decrease in temperature) ΔU is negative.

Application of First Law of Thermodynamics to Different Chemical Processes:

Isothermal Process:

Internal energy is a function of temperature. As the temperature is constant, the internal energy is also constant. Hence there is no change in internal energy.     ΔU = 0

By the first law of thermodynamics,

ΔU = q   + W

∴ 0 = q   + W

∴ q = -W or W = – q

Thus in the isothermal process heat absorbed is entirely used for doing work on the surroundings or the work done by the surrounding at constant pressure results in the release of the heat (energy) by the system.

Adiabatic Process:

In an adiabatic process, there is no exchange of heat      q = 0

By the first law of thermodynamics

ΔU = q   + W

ΔU = 0   + W

∴   W = q

Thus the increase in internal energy of a system is due to work done by the surroundings on the system or work done by the system on the surroundings is due to the expense of internal energy of the system.

Isochoric Process:

In isochoric process there is no change in volume   ΔV = 0, Thus the work done W = P Δ V = 0

By the first law of thermodynamics,

ΔU = q   + W

∴ ΔU = q   + 0

∴ ΔU = q

Thus the increase in internal energy of a system is due to the absorption of the heat from the surroundings or the decrease in internal energy of a system is due to the release of the heat from the system to the surroundings.

Isobaric Process:

In isobaric process, there is no change in pressure   ΔP = 0 Thus work done W = – P_ext ΔV

By the first law of thermodynamics,

ΔU = q   + W

∴ ΔU = q   – P_ext ΔV

Thus the increase in internal energy of a system is due to the absorption of the heat from the surroundings or the decrease in internal energy of a system is due to release of the heat from the system to the surroundings.

Note:

Most of the chemical reactions take place at constant pressure.

Second Law of Thermodynamics:

We observe that heat flows from the hot end of a metal rod flows to its cold end. The opposite flow of heat is not taking place. The first law (law of conservation of energy) allows heat flow from cold end to hot end. It is possible when heat lost by the cold end is equal to the heat gained by hot end. Thus energy is conserved. But such heat transfer is not possible. Thus the first law of thermodynamics is insufficient to put a restriction on the direction of the heat flow.  To overcome the deficiency of the first law, the second law of thermodynamics is proposed using human experience.

Statement:

Heat cannot be completely converted into an equivalent amount of work without producing permanent changes either in the system or its surroundings. or The spontaneous flow of heat is always unidirectional, from higher temperature to lower temperature.

Previous Topic: Concept of Maximum Work

Next Topic: Enthalpy of a System

Science > Chemistry > Chemical Thermodynamics and Energetics > First Law of Thermodynamics

The post Laws of Thermodynamics appeared first on The Fact Factor.

According to the first law of thermodynamics, ΔU = q  +  W    In an isothermal process, ΔU  =   0, ∴   q   =   – W

Therefore, all the heat absorbed by the system is utilized to do work. (i.e. maximum utilization of the energy takes place. )

The work of expansion is given by the product of external pressure and the volume change. W = – P_ext ΔV

In any expansion, the external pressure must be less than the pressure of the gas. If the external pressure is zero, the work done is also zero as the gas expands into the vacuum. If the external pressure is increased gradually, more and more work will be done by the gas during expansion. If the external pressure becomes equal to the pressure of the gas, there will be no change in the volume and thus ΔV = 0. The work done is also zero. If P_ext is more than the pressure of the gas cannot expand. Therefore, when Pext becomes P then ΔV will be maximum. In the reversible process, P_ext is always less than the pressure of the gas, by an infinitesimally small quantity.

W =  (P –  dp) dV

In the equation W tends to the maximum as (P – dp) tends to P or dp tends to zero. Therefore work done in an isothermal reversible expansion of an ideal gas is maximum work.

Conditions for Maximum Work:

All the changes taking place in a system during the process are reversible. All the changes taking place in a system during the process should take place in infinitesimally small infinite steps. During the change, the driving and opposing forces should defer by an infinitesimally small amount. The system remains in mechanical equilibrium with the surroundings.

Expression for Maximum Work:

Work done in an isothermal reversible expansion is maximum work.

Consider  ‘n ‘ moles of an ideal gas enclosed in a cylinder fitted with a weightless, frictionless, airtight movable piston. Let the pressure of the gas be P which is equal to external atmospheric pressure P. Let the external pressure be reduced by an infinitesimally small amount dp and the corresponding small increase in volume be dV. So the small work done in the expansion process

dW  =  – P_ext . dV

∴ dW =  – (P  –  dp). dV

∴ dW =   – (P.dV  –  dp.dV)  …..  (1)

Since both dp and dV are very small, the product dp.dV is very small and can be neglected in comparison with P.dV. Then the above equation becomes

dW  =   – P.dV                ……….  (2)

When the expansion of the gas is carried out reversibly then there will be series of such p.dV terms. The total maximum work W_max can be obtained by integrating above equation between the limits V₁ to  V₂. Where V₁ is initial volume and V₂ is final volume.

For an isothermal expansion, Boyle’s law is applicable.
Hence P₁V₁ = P₂V₂ i.e. V₂ / V₁ = P₁ / P₂

Where P₁  and P₂  are initial and final pressure respectively.

Notes:

• As the ratio of volumes or pressure is used in the above equation when work done is to be calculated the volumes and the pressures may be expressed in any unit, provided both the quantities are expressed in the same unit.
• The work obtained using the above equation will be in joule if R is taken in the S.I. unit.
• The number of moles  = Wt. in gram / Molecular wt.in grams
• Absolute temperature T k   = to C   + 273

Work is Path Function:

The work done in isothermal constant pressure process is given by

W   =    – Pext ΔV i.e. W   =    – Pext  ( V₂  –  V₁) 

Where, Pext = External opposing pressure

V₁ = Initial volume V₂ = Final volume.

The work done in the isothermal reversible process is given by

Where, n = Number of moles of the gas R = Universal gas constant

T = Absolute temperature of the gas

V₁ = Initial volume V₂ = Final volume

We can see that the work done depends on the manner or the conditions under which the process carried out. Thus it is not dependent on initial and final conditions of the system but on the path followed by the system. Hence work is path function or it is not a state function.

Numerical Problems on Maximum Work:

Example – 01:

3 moles of an ideal gas are expanded isothermally and reversibly from volume of 10 m³ to the volume 20 m³ at 300 K. Calculate the work done.

Given: n = 3 moles, V₁ = 10 m³, V₂ = 20 m³, T = 300 K, R = 8.314 J K^-1 mol^-1.

To Find: Work done =?

Solution:

Work done in isothermal reversible process is given by

W_max = -2.303 nRT log₁₀(V₂/V₁)

∴ W_max = -2.303 × 3 mol × 8.314 J K^-1 mol^-1 × 300 K × log₁₀( 20 m³/ 10 m³)

∴ W_max = -2.303 × 3 × 8.314 × 300 × log₁₀(2)

∴ W_max = -2.303 × 3 × 8.314 × 300 × 0.3010

∴ W_max = – 5187 J = – 5.187 kJ

Ans: maximum work done =  – 5187 J = – 5.187 kJ

Negative sign indicates the work is done by the system on the surroundings

Example – 02:

24 g of oxygen are expanded isothermally and reversibly from 1.6 × 10⁵ Pa pressure to 100 kPa at 298 K. Calculate the work done.

Given: m = 24 g, P₁ = 1.6 × 10⁵ Pa, P₂ = 100 kPa = 100 × 10³ Pa = 1 × 10⁵ Pa, T = 298 K, R = 8.314 J K^-1 mol^-1.

To Find: Work done =?

Solution:

Number of moles of oxygen = Given mass of oxygen / molecular mass of oxygen

Number of moles of oxygen = 24 g / 32 g mol^-1 = 0.75 mol

Work done in isothermal reversible process is given by

W_max =  -2.303 nRT log₁₀(P₁/P₂)

∴ W_max = -2.303 × 0.75 mol × 8.314 J K^-1 mol^-1 × 298 K × log₁₀( 1.6 × 10⁵ Pa / 1 × 10⁵ Pa)

∴ W_max = -2.303 × 0.75 × 8.314× 298 × log₁₀( 1.6)

∴ W_max = -2.303 × 0.75 × 8.314 × 298  × 0.2041

∴ W_max = – 873.4 J

Ans: maximum work done =  – 873.4 J

Negative sign indicates the work is done by the system on the surroundings

Example – 03:

4.4 × 10^-2 kg of CO₂ is compressed isothermally and reversibly at 293 K from the initial pressure of 150 kPa when work obtained is 1.245 kJ. Find the final pressure.

Given: m = 4.4 × 10^-2 kg, P₁ = 150 kPa, Maximum work of compression = W = + 1.245 kJ = 1245 J, T = 293 K, R = 8.314 J K^-1 mol^-1, P₂ =?

To Find: Final pressure =?

Solution:

Number of moles of CO₂ = Given mass of CO₂ / molecular mass of CO₂

Number of moles of CO₂ = n = 4.4 × 10^-2 kg / 44 × 10^-3 kg mol^-1 = 1 mol

Work done in isothermal reversible process is given by

W_max = -2.303 nRT log₁₀(P₁/P₂)

∴ 1245 J = -2.303 × 1 mol × 8.314 J K^-1 mol^-1 × 293 K × log₁₀(P₁/P₂)

∴ 1245 J = -2.303 × 1 mol × 8.314 J K^-1 mol^-1 × 293 K × log₁₀(P₂/P₁)

∴ log₁₀(P₂/P₁) = (1245) J / (2.303 × 1 × 8.314 × 293)  J

∴ log₁₀(P₂/P₁) = 0.2219

∴ (P₂/P₁) = Antilog (0.2219) = 1.667

∴ P₂ = 1.667 ×  P₁  = 1.667 × 150 kPa = 250 kPa

Ans: Final Pressure is 250 kPa

Example – 04:

2.8 × 10^-2 kg of nitrogen is expanded isothermally and reversibly at 300 K from the initial pressure of 15.15 × 10⁵ Nm^-2 when work obtained is 17.33 kJ. Find the final pressure.

Given: m = 42.8 × 10^-2 kg, P₁ = 15.15 × 10⁵ Nm^-2, Maximum work of expansion = W = – 17.33 kJ = – 17.33 × 10³ J, T = 293 K, R = 8.314 J K^-1 mol^-1, P₂ =?

To Find: Final Pressure =?

Solution:

Number of moles of Nitrogen = Given mass of Nitrogen / molecular mass of Nitrogen

Number of moles of Nitrogen = n = 2.8 × 10^-2 kg / 28 kg mol^-1 = 1 mol

Work done in isothermal reversible process is given by

W_max = -2.303 nRT log₁₀(P₁/P₂)

∴ – 17.33 × 10³ J = -2.303 × 1 mol × 8.314 J K^-1 mol^-1 × 300 K × log₁₀(P₁/P₂)

∴ 17.33 × 10³ J = 2.303 × 1 mol × 8.314 J K^-1 mol^-1 × 300 K × log₁₀(P₁/P₂)

∴ log₁₀(P₁/P₂) = (17.33 × 10³) J / (2.303 × 1 × 8.314 × 300)  J

∴  log₁₀(P₁/P₂) = 3.0170

∴ (P₁/P₂) = Antilog (3.0170) = 1040

∴ P₂ =  P₁ / 1040 = 15.15 × 10⁵ Nm^-2 / 1040 =  14056.7 Nm^-2

Ans: Final Pressure is 14056.7 Nm^-2

Example – 05:

3 moles of an ideal gas are compressed isothermally and reversibly at 22° C to a volume 2L when work done is 2.983 kJ. Find the initial volume.

Given: n = 3 mol, V₂ = 2 L, Maximum work of compression = W = + 2.983 kJ = 2.983 × 10³ J , T = 22° C = 22 + 273 = 295 K, R = 8.314 J K^-1 mol^-1, V₁ = ?,

To Find: Initial volume =?

Solution:

Work done in isothermal reversible process is given by

W_max =  -2.303 nRT log₁₀(V₂/V₁)

∴ 2.983 × 10³ J =  -2.303 ×3 mol × 8.314 J K^-1 mol^-1 × 295 K × log₁₀(V₂/V₁)

∴ 2.983 × 10³ J = 2.303 ×3 mol × 8.314 J K^-1 mol^-1 × 295 K × log₁₀(V₁/V₂)

∴ log₁₀(V₁/V₂) = 2.983 × 10³ J / (2.303 × 3 × 8.314 × 295)  J

∴ log₁₀(V₁/V₂) = 0.1760

∴ (V₁/V₂) = Antilog (0.1760) = 1.5

∴ V₁= 1.5  × V₂  = 1.5  ×  2 L  = 4 L

Ans: Initial Volume = 4 L

Example – 06:

280 mmol of an ideal gas occupy 12.7 L at 310 K. Calculate the work done when gas expands a) isothermally against constant external pressure 0.25 atm, b) isothermally and reversibly c) in a vacuum until its volume becomes 3.3 L.

Given: n = 280 mmoles = 0.280 mol, V₁ = 12.7 L, ΔV = 3.3 L, T = 310 K, R = 8.314 J K^-1 mol^-1. External Pressure = 0.25 atm

To Find: Work done =?

Solution:

a)  The gas expands isothermally against constant external pressure 0.25 atm

W = – P_ext × ΔV = – 0.25 atm × 3.3 L = – 0.825 L atm

W = – 0.825 L atm × 101.3  J L^-1 atm^-1 = – 83.6 J

Hence work done = – 83.6 J

Negative sign indicates the work is done by the system on the surroundings

b) Gas expands isothermally and reversibly

ΔV = (V₂ – V₁) = (V₂ –  12.7 L) =  3.3 L

V₂ = 3.3 L + 12.7 L = 16 L

Work done in isothermal reversible process is given by

W_max = -2.303 nRT log₁₀(V₂/V₁)

∴ W_max = -2.303 × 0.280 mol × 8.314 J K^-1 mol^-1 × 310 K × log₁₀( 16 L/ 12.7 L)

∴ W_max = -2.303 × 0.280 × 8.314 × 310 × log₁₀(1.260)

∴ W_max = -2.303 × 0.280 × 8.314 × 310 × 0.1004

∴ W_max = – 166.9  J

Maximum work = 166.9 J

Negative sign indicates the work is done by the system on the surroundings

c) Gas expands in vacuum

In this case, it is a free expansion, there is no opposing pressure. Hence P_ext = 0

W = – P_ext × ΔV = 0 atm × 3.3 L  = 0

Work done = 0

Example – 07:

 300 mmol of a perfect gas occupies 13 L at 320 K. Calculate the work done in joules when the gas expands a) isothermally against the constant external pressure of 0.20 atm. b) Isothermal and reversible process c) Into vacuum until the volume of the gas is increased by 3 L.

Given: n = 300 mmoles = 0.3 mol, V₁ = 13 L, ΔV = 3 L, T = 320 K, R = 8.314 J K^-1 mol^-1. External Pressure = 0.20 atm

To Find: Work done =?

Solution:

a)  The gas expands isothermally against constant external pressure 0.20 atm

W = – P_ext × ΔV = – 0.20 atm × 3 L = – 0.6 L atm

W = – 0.6 L atm × 101.3 J L^-1 atm^-1 = – 60.78 J

Hence work done = – 60.78 J

Negative sign indicates the work is done by the system on the surroundings

b) Gas expands isothermally and reversibly

ΔV = (V₂ –  V₁)  =  (V₂ –  13 L) =  3 L

V₂ = 3 L + 13 L = 16 L

Work done in isothermal reversible process is given by

W_max = -2.303 nRT log₁₀(V₂/V₁)

∴ W_max = -2.303 × 0.3 mol × 8.314 J K^-1 mol^-1 × 320 K × log₁₀( 16 L/ 13 L)

∴ W_max = -2.303 × 0.3 × 8.314 × 320 × log₁₀(1.231)

∴ W_max = -2.303 × 0.3 × 8.314 × 320 × 0.0902

∴ W_max = – 165.8  J

Maximum work = – 165.8  J

Negative sign indicates the work is done by the system on the surroundings

c) Gas expands in vacuum

In this case, it is the free expansion, there is no opposing pressure. Hence P_ext = 0

W = – P_ext × ΔV = 0 atm × 3 L  = 0

Work done = 0

Previous Topic: Pressure-Volume Type Work

Next Topic: First Law of Thermodynamics

Science > Chemistry > Chemical Thermodynamics and Energetics > Concept of Maximum Work

The post Concept of Maximum Work appeared first on The Fact Factor.

Pressure Volume Work:

Consider an ideal gas having definite mass (say n moles) be enclosed in a cylinder fitted with weightless, frictionless, tightly fitted, movable piston. Let ‘A’ be the area of the cross-section of the cylinder. Let the gas expand from volume V₁ to V₂ against constant external pressure P_ext which is exerted on the piston. Due to expansion, the piston moves upward by a distance ‘d’. In this process, the system loses energy to the surroundings. Hence the work done by the system is negative.

Work done = – Opposing Force x displacement

∴   W = – F x displacement    …….  (1)

Now, pressure is force per unit area.

∴ Pressure   = Force /  Area

∴ Force = Pressure x  Area

∴       F        =      P   x    A   ……..  (2)

Substituting this value in equation (1)

∴   W   =     –  P   x   A    x   dx     ………. (3)

But,  A   x  dx  = Volume through gas expands

=   Change in volume =  ( V₂  –  V₁) = Δ V

Substituting this value in equation (3)

W = – P_ext ΔV

This is an expression for pressure volume work done in terms of pressure and volume in the isothermal expansion of an ideal gas against constant external pressure.

Notes:

• It is the external pressure against gas expands i.e. P_ext and not the pressure of the gas that is used in evaluating the pressure volume work done.
• The final equation shows that work  W  depends only on the change in volume (ΔV) and opposing pressure (P_ext) and not on the quantity of the gas (i.e. it is independent of the number of moles) and Temperature of the gas (T).
• The work done by the system in a process depends on the way or manner in which it is carried out. Work, therefore, is not a state function. It is a path function.

Free Expansion:

The free expansion is also known as work of expansion in a vacuum. When a gas expands in a vacuum there is no opposing force i.e.  P_ext = 0.

For getting work done opposing force is necessary, hence no work is done when a gas expands in a vacuum. Such an expansion is called as free expansion.

W   =    – P_ext ΔV

∴ W   =  0 × ΔV = 0

Work Done in a Cyclic Process:

A cyclic process is one which consists of a series of intermediate steps, at the end of which the system returns to its initial state.

Since in a cycle, initial and final states are the same.

∴ Δ(state function)   =   0

∴ ΔU =  0 (i.e. No change in internal energy)

Now,   ΔU   = q + W

∴ 0 = q  +   W

∴ q = -W

Thus in a cyclic process heat absorbed by the system from the surrounding (q) is equal to work done (W) by the system on the surrounding.

Difference Between Heat and Work: 

When heat is added to gas the random motion of molecules of gas increases, thus heat is a stimulus to gas which increases the random motion of the gas molecules. Thus heat is a random form of energy. Now let us consider work done on a system by compressing the gas in a cylinder. The movement of the piston makes the gas molecules to move in the direction of applied force. Thus work is a stimulus which increases the organized motion of the molecules of the gas. Thus work is an organized form of energy.

Sign Convention Used in Thermodynamics:

• During the expansion, V₂ > V₁, Hence ΔV is positive.    Hence work done W is negative. Thus when the work is done by the system on the surrounding i.e. Work of expansion is taken as negative (- W).
• During compression, V₂ < V₁, Hence ΔV is negative. Hence work done W is positive. Thus when the work done on the system by the surrounding i.e. Work of compression is taken as positive (+W).

Numerical Problems on Pressure Volume Work:

Example – 01:

2 moles of an ideal gas are expanded isothermally from volume of 15.5 L to the volume 20 L against a constant external pressure of 1 atm. Calculate the pressure-volume work in L atm and J.

Given: n = 2 moles, V₁ = 15.5 L, V₂ = 20 L, P_ext = 1 atm

To Find: Work Done =?

Solution:

Work done in an isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 1 atm × (20 L – 15.5 L)

∴ W = – 1 atm × (4.5 L) = – 4.5 L atm

W = – 4.5 L atm × 101.3 J L^-1 atm^-1 = – 455.8 J

Ans: work done = – 4.5 L atm or – 455.8 J

The negative sign indicates the work is done by the system on the surroundings

Example – 02:

2 moles of an ideal gas are compressed isothermally from volume of 10 dm³ to the volume 2 dm³ against a constant external pressure of 1.01 × 10⁵ Nm^-2. Calculate the pressure-volume work done.

Given: n = 2 moles, V₁ = 10 dm³ = 10 × 10^-3 m³, V₂ = 2 dm³ = 2 × 10^-3 m³, P_ext = 1 atm.

To Find: Work Done =?

Solution:

Work done in an isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 1.01 × 10⁵ Nm^-2 × (2 × 10^-3 m³ – 10 × 10^-3 m³)

∴ W = – P_ext × (V₂ – V₁) = – 1.01 × 10⁵ Nm^-2 × (- 8 × 10^-3 m³)

∴ W = + 8.08 × 10² J = + 808 J

Ans: work done = + 808 J

Positive sign indicates the work is done by the surrounding on the system

Example – 03:

3 moles of an ideal gas are expanded isothermally from volume of 300 cm³ to the volume 2.5 L against a constant external pressure of 1.9 atm at 300 K. Calculate the pressure volume work in L atm and J.

Given: n = 3 moles, V₁ = 300 cm³= 0.3 L, V₂ = 2.5 L, P_ext = 1.9 atm

To Find: Work Done =?

Solution:

Work done in an isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 1.9 atm × (2.5 L – 0.3 L)

∴ W = – 1.9 atm × (2.2 L) = – 4.18 L atm

W = – 4.18 L atm × 101.3 J L^-1 atm^-1 = – 423.4 J

Ans: work done = – 4.5 L atm or – 423.4 J

The negative sign indicates the work is done by the system on the surroundings

Example – 04:

1 mole of an ideal gas is compressed isothermally from a volume of 500 cm³ against a constant external pressure of 1.216 × 10⁵ Pa. The pressure-volume work involved in the process is 36.5 J. calculate the final volume.

Given: n = 1 mole, V₁ = 500 cm³= 500 × 10^-6 m³, P_ext = 1.216 × 10⁵ Pa = = 1.216 × 10⁵ Nm^-2, Work of compression = + 36. 5 J, V₂ =?

To Find: Final volume = V₂ =?

Solution:

Work done in isothermal process is given by W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁)

∴ 36.5 J= – 1.216 × 10⁵ Nm^-2 × (V₂ – 500 × 10^-6 m³)

∴ 36.5 J/ 1.216 × 10⁵ Nm^-2 = – (V₂ – 500 × 10^-6 m³)

∴300 × 10^-6 m³ = (500 × 10^-6 m³ – V₂)

∴    V₂    = (500 × 10^-6 m³ – 300 × 10^-6 m³)

∴    V₂    = 200 × 10^-6 m³ = 200 cm³

Ans: final volume = 200 cm³

Example – 05:

1 mole of an ideal gas is compressed isothermally from volume of 20 L to 8 L against constant external pressure, when pressure volume work obtained is 44.9 L atm. Find the constant external pressure.

Given: n = 1 mol, V₁ = 20 L, V₂ = 8 L, Work of compression W = + 44.9 L atm, P_ext =?

To Find: External pressure =?

Solution:

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁)

∴ 44.9 L atm = – P_ext × (8 L – 20 L)

∴ 44.9 L atm = – P_ext × (- 12 L)

∴ P_ext   =44.9 L atm / 12 L)

∴ P_ext   = 3.74 atm

Ans: constant external pressure is 3.74 atm

Example – 06:

One mole of a gas expands by 3 L against a constant pressure of 3 atmosphere. Calculate the pressure volume work done in a) L-atm b) joules and c) calories.

Given: n = 1 mole, Δ V = 3 L, P_ext = 3 atm

To Find: Work Done =?

Solution:

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – 3 atm × 3 L = – 9 L atm

∴ W = – 9 L atm × 101.3 J L^-1 atm^-1 = – 911.7 J

∴ W = – 911.7 J / 4.184 J cal^-1 = – 217.9 cal

Ans: work done = – 9 L atm or – 911.7 J  or -217.9 cal

Negative sign indicates the work is done by the system on the surroundings

Example – 07:

100 mL of ethylene(g) and 100 mL of HCl (g) are allowed to react at  2 atm pressure as per the reaction given below. Calculate pressure volume type of work in it in joules.

C₂H_4(g)    +  HCl_(g)    →     C₂H₅Cl_(g)

Solution: 

Given: P_ext = 2 atm and the given reaction is

C₂H_4(g)    +  HCl_(g)    →     C₂H₅Cl_(g)

1 Vol               1Vol                      1Vol

100 mL            100 mL                  100 mL

Thus 100 ml of C₂H_4(g) reacts with 100 mL of HCl_(g)  to give 100 mL of C₂H₅Cl_(g).

Initial volume = Volume of reactants = 100 mL + 100 mL = 200 mL = 0.2 L

Final volume = Volume of products = 100 mL = 0.1 L

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 2 atm × (0.1 L – 0.2 L)

∴ W = – 2 atm × (- 0.1 L) = 0.2  L atm

W = 0.2 L atm × 101.3 J L^-1 atm^-1 = 20.26 J

Ans: work done = + 20.26 J

Positive sign indicates the work is done by the surroundings on the system

Example – 08:

A gas cylinder of 5 L capacity containing 4 kg of helium gas at 27 °C developed a leakage leading to the escape of the gas into atmosphere. If atmospheric pressure is 1.0 atm. calculate the pressure volume work done by the gas assuming ideal behaviour.

Given: V₁ = 5 L, mass of gas m = 4 kg = 4 × 10³ g, T = 27 °C = 27 +273 = 300 K,  P_ext = P = 1.0 atm, R = 0.0821 L atm K^-1 mol^-1.

To Find: Work Done =?

Solution:

Number of moles = Given mass of He / Molecular mass of He = 4 × 10³ g / 4 g = 10³

By ideal gas equation, we have P V₂ = nRT

∴   V₂ = nRT / P = 10³ × 0.0821 × 300 / 1   = 2.463 × 10⁴ L = 24630 L

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 1 atm × (24630 L – 5 L)

∴ W = – 1 atm × (24625 L) = – 24625 L atm

W = – 24625 L atm × 101.3 J L^-1 atm^-1 = – 2.494 × 10⁶ J = – 2494 kJ

Ans: work done = – 2494 kJ

Negative sign indicates the work is done by the system on the surroundings

Example – 09:

1.6 mol of water evaporates at 373 K against atmospheric pressure of 1 atm. Assuming ideal behaviour of water vapours calculate the work done.

Given: n = 1.6 mole, T = 373 K, P_ext = P = 1.0 atm, R = 0.0821 L atm K^-1 mol^-1.

To Find: Work Done =?

Solution:

The molecular mass of water 18 g. and the density of water is 1 g per cc

Hence initial volume of water = V₁ = 18 x 1.6 x 10^-3 L = 0.0288 L

By ideal gas equation, we have P V₂ = nRT

∴   V₂ = nRT / P = 1.6 × 0.0821 × 373 / 1   = 48.93 L

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 1 atm × (48.93 L – 0.0288 L)

∴ W = – 1 atm × (48.90 L) = – 48.90  L atm

W = – 48.90 L atm × 101.3 J L^-1 atm^-1 = – 4954.8 J

Hence work done = – 4954.8 J

Negative sign indicates the work is done by the system on the surroundings

Example – 10:

1.0 mol of water evaporates at 373 K against atmospheric pressure of 749.8 mm of Hg. Assuming ideal behaviour of water vapours calculate the pressure volume work done.

Given: n = 1.0 mole, T = 373 K, P_ext = P = 749.8 mm of H = 749.8 / 760 = 0.987 atm, R = 0.0821 L atm K^-1 mol^-1.

To Find: Work Done =?

Solution:

The molecular mass of water 18 g. and the density of water is 1 g per cc

Hence initial volume of water = V₁ = 18 x 1.0 x 10^-3 L = 0.018 L

By ideal gas equation, we have P V₂ = nRT

∴   V₂ = nRT / P = 1.0 × 0.0821 × 373 / 0.987   = 31.03 L

Work done in isothermal process is given by   W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 0.987 atm × (31.03 L – 0.018 L)

∴ W =  –  0.987 atm × (31.01 L) = – 30.61  L atm

W = – 30.61 L atm × 101.3 J L^-1 atm^-1 = – 3101 J

Ans: work done = – 3101 J

Negative sign indicates the work is done by the system on the surroundings

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In this article, we shall study Kirchhoff’s laws, construction, working and uses of Wheatstone’s metre bridge. There are two Kirchhoff’s laws viz: current law or junction law and voltage law or mesh law.

Kirchhoff’s Current Law or Kirchhoff’s Junction Law:

Statement: 

In an electric network, the algebraic sum of currents at any junction is zero.

Explanation:

Any point in the circuit where the current split is called a junction. Currents approaching junction are taken positive and currents going away from junction are taken as negative. Let us consider a junction of a circuit as shown in the figure.

Applying Kirchhoff’s junction law at point O.

I₁ + I₂ – I₃ – I₄ – I₅ = 0

Since there is no loss or gain of the charge at the junctions, this law is in accordance with the law of conservation of charge.

Kirchhoff’s Voltage Law OR Kirchhoff’s Mesh Law:

Statement: 

The algebraic sum of the products of the current and resistance of each part of a closed circuit (mesh or loop) is equal to the algebraic sum of the e.m.f.s in that closed circuit.

Sign Convention:

• When passing through the circuit in the direction of current there is drop in the potential hence the potential difference should be taken as negative.
• When passing through the circuit in the opposite direction of current there is an increase in the potential hence the potential difference should be taken as positive.
• When passing through a cell from the negative terminal to the positive terminal the e.m.f. of a cell is taken as positive.
• When passing through a cell from the positive terminal to the negative terminal the e.m.f. of a cell is taken as negative.

Explanation:

Consider the following circuit

Let us consider a closed-loop E₁ARBr₁

E₁ – (I₁ + I₂) R – I₁r₁ = 0

Let us consider a closed-loop E2ARBr2

E2 – (I₁ + I2) R – I₂r₂ = 0

Let us consider a closed-loop E₁AE₂BE₁

E₁ – E₂ + I₂r₂ – I₁r₁ = 0

Wheatstone’s Network (Application of Kirchhoff’s Laws):

Circuit Arrangement:

A Wheatstone’s network consists of four resistances connected such that they form a quadrilateral of resistances. A battery is connected (or p.d. is applied) between one pair of opposite corners of the quadrilateral. A galvanometer is connected between another pair of opposite corners.

When the values of the resistance are such that, the galvanometer shows no deflection or null deflection then the network is said to be balanced Wheatstone’s network.

Condition for Balanced Wheatstone’s Network:

Let R₁, R₂, R₃, and R₄ be the four resistances connected as shown in the diagram to form Wheatstone’s network as shown in the diagram. Let I₁, I₂ be the currents through the resistances R₁ and R₃ respectively. Let I_G be the current through the galvanometer whose resistance is G. In balanced condition the current through galvanometer is zero hence I_G = 0

Applying Kirchhoff’s voltage law to the loop ABDA,

I₁R₁ –I_GG + I₂R₃ = 0

∴ – I₁R₁ –(0) G +I₂R₃ = 0

∴ – I₁R₁ + I₂R₃ = 0

∴ I₁R₁ = I₂R₃ ……….. (1)

Applying Kirchhoff’s voltage law to the loop BCDB

– (I₁ – I_G) R₂ + (I₂ + I_G) R₄   + I_GG = 0

∴  – (I₁ – 0)R₂ + (I₂ +0)R4   + (0)G = 0

∴ – I₁ R₂ + I₂R4  = 0

I₁ R₂ = I₂R4 ……..   (2)

Dividing equation (1) by (2)

This is the required condition for balanced Wheatstone’s network.

Condition for Balanced Wheatstone’s network using Ohm’s law:

When the values of the resistance are such that, the galvanometer shows no deflection or null deflection then the network is said to be balanced Wheatstone’s network. Thus potential at point B is equal to the potential at D. Thus (V_B = V_D)

∴ V_A – V_B = V_A – V_D

∴ By Ohm’s law I₁R₁ = I₂R₃ ……… (1)

Similarly V_B – V_C = V_D – V_C

∴ By Ohm’s law (I₁ – I_G)R₂ = (I₂ + I_G)R₄ ……… (2)

But no current flows through the galvanometer (I_G = 0)

I₁R₂ = I₂R₄ ……… (2)

Dividing equation (1) by (2)

This is the required condition for balanced Wheatstone’s network.

Wheatstone’s Metre Bridge:

The value of an unknown resistance can be determined by using Wheatstone’s meter bridge.

Construction:

It consists of a uniform wire AC, one meter long, stretched on a wooden board.  The two ends of the wire are fixed to two thick L shaped copper strips.  The third thick and straight copper strip is so put that it forms two gaps with the two copper strips.

The unknown resistance X is connected in one gap and a resistance box R (known resistance) is connected in the other gap. The junction of X and R is connected to one terminal of a galvanometer G.  The other terminal of the galvanometer is connected to a pencil Jockey which can slide along the wire AC. A cell, key, and rheostat are connected in series with the wire.

Working:

A suitable resistance R is taken in the resistance box and the current is sent around the circuit by closing the key K. The jockey is touched to different points of the wire AC and a point of contact D for which the galvanometer shows zero deflection is found. As at point D the galvanometer is showing null deflection, point D is called the null point.

Let the distance of the point from A be l₁ and that from C is l₂. These two distances are measured. Let σ be the resistance per unit length of wire. As the network is balanced

Using this formula unknown resistance X can be calculated.

Possible Errors in Metre Bridge Experiment:

• If the wire is not uniform, the resistance per unit length of wire is not the same throughout the wire and thus there is a possible error in the determination of unknown resistance.
• There may be the error due to contact resistances of the points where the wires are connected to copper strips.

Minimization of Errors:

• The resistance X and R are interchanged and experiment is repeated.
• Value of R is so selected that the deflection in the galvanometer is at the centre of the scale.
• The average of the readings of a number of readings gives most probable value.

Kelvin’s Method to Find Resistance of a Galvanometer:

By this method, the resistance of galvanometer itself can be found.

Circuit Arrangement:

The galvanometer whose resistance to be found is connected in one gap and the known resistance R is connected in another gap. The galvanometer itself is used to determine the condition of equilibrium without locating null point. The junction of galvanometer G and known resistance R is connected to the jockey.

The circuit is closed and adjusting the rheostat a suitable current is passed through the circuit. A suitable resistance is taken out from the resistance of the box and deflection in the galvanometer is noted. A jockey is moved over wire AC. A point D is noted for which the galvanometer shows the same deflection as before. Thus D point is such that at this point the deflection is the same with or without the jockey. Thus D is equal deflection point.

Distances of point D from two ends are measured.  Let AD = l₁ and CD =l₂. If G is the resistance of the galvanometer, then using the following formula the value of G can be calculated.

Previous Topic: Temperature Dependence of Resistance

Next Topic: Numerical Problems on Metre Bridge

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