In this article, we shall study to find r.ms. speed of gas molecules, density of gas, pressure exerted by the gas using kinetic theory of gases.

Example 01:

Calculate the R.M.S. velocity of Hydrogen molecules at N.T.P. given: Density of Hydrogen at N.T.P. = ρ = 8.957 x 10^-2 kg/m³. Density of mercury = 13600 kg/m³, g = 9.8 m/s².

Given: Density of hydrogen = ρ = 8.957 x 10^-2 kg/m³, condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9. 8 N/m², Density of Mercury = 13600 kg/m³, g = 9.8 m/s².

To Find: r.m.s. speed = C =?

Solution:

Ans: r.m.s. velocity of hydrogen molecule is 1842 m/s 0r 1.842 km/s

Example 02:

Find the R.M.S. velocity of Nitrogen molecules at N.T.P. given that the density of Nitrogen at this temperature is 1.25 g/litre.

Given: Density = ρ = 1.25 g/litre = 1.25 x 1 = 1.25 kg/m³, condition N.T.P., P = 1.013 x 10⁵ N/m².

To Find: r.m.s. speed = C =?

Solution:

Ans: r.m.s. velocity of nitrogen molecule is 493.1 m/s

Example 03:

Calculate the R.M.S. velocity of Oxygen molecules at a pressure of 1.013 x 10⁵ N/m² (N.T.P.) given the density of Oxygen is 1.44 kg/m³.

Given: Density = ρ = 1.44 kg/m³, P = 1.013 x 10⁵ N/m².

To Find: r.m.s. speed = C =?

Solution:

Ans: r.m.s. velocity of oxygen molecule is 459.4 m/s

Example 04:

Calculate the density of He at N.T.P. given that R.M.S. velocity of He molecules at N.T.P. is 1300 m/s.

Given: Condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², Density r.m.s. speed = C = 1300 m/s.

To Find: Density = ρ =?

Solution:

Ans: density of He is 0.1798 kg/m³

Example 05:

Determine the pressure of oxygen at 0 °C if the density of oxygen at NTP is 1.44 kg/m³ and r.m.s speed of the molecule at NTP is 456.4 m/s.

Given: Density = ρ = 1.44 kg/m³, condition N.T.P., P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², r.m.s. speed = C = 456.4 m/s

To Find: Pressure of gas = P =?

Solution:

Ans: Pressure of oxygen = 10⁵ N/m²

Example 06:

Two gases are at temperatures of 77 ^oC and 27 ^oC. What is the ratio of the R.M.S. velocities of the molecules of the two gases?

Given: temperature of first gas = T₁ = 77 ^oC = 77 + 273 = 350 K, temperature of second gas = T₂ = 27 ^oC = 27 + 273 = 300 K

To Find: Ratio of r.m.s speeds =?

Solution:

Ans: The ratio of r.m.s. speed is 1.08:1

Example 07:

At what temperature will the R.M.S. speed of the molecules of gas be three times its value at N.T.P.?

Given: Condition = N.T.P., T₁ = 273 K, C₂ = 3C₁

To Find: Temperature = T₂ =?

Solution:

Ans: At a temperature of 2184 °C the r.m.s. speed of the molecules of a gas is three times its value at N.T.P.

Example 08:

At what temperature will the R.M.S. speed of the molecules of gas be four times its value at N.T.P.?

Given: Condition = N.T.P., T₁ = 273 K, C₂ = 4C₁

To Find: Temperature = T₂ =?

Solution:

Ans: At a temperature of 4095 °C the r.m.s. speed of the molecules of a gas is four times its value at N.T.P.

Example 09:

The R.M.S. velocity of Nitrogen molecules at N.T.P. is 497 m/s. Calculate the R.M.S. velocity of Hydrogen molecules at N.T.P. At what temperature will the R.M.S. velocity of Nitrogen molecules be 994 m/s?

Part – I:

Given: For nitrogen: Condition = N.T.P., T_N = 273 K, C_N = 497 m/s, P_N = 1.013 x 10⁵ N/m², for oxygen: Condition = N.T.P., T_N = 273 K, P_H = 1.013 x 10⁵ N/m².

To Find: r.m.s. speed of hydrogen = C_H =?

Solution:

Part – II:

Given: Condition = N.T.P., T₁ = 273 K, C₁ = 497 m/s, C₂= 994 m/s

To Find: Temperature = T₂ =?

Solution:

Ans: r.m.s. speed of hydrogen molecules at NTP1860 m/s, at a temperature of 819oC the speed of nitrogen molecules at NTP 994 m/s,

Example 10:

Calculate the R.M.S. velocity of Oxygen molecules at 27 °C. The density of Oxygen at N.T.P. 1.44 kg/m³.

Given: For nitrogen: Condition = N.T.P. T₁ = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², density of oxygen = ρ = 1.44 kg/m³, Temperature = T₂ = 27 ^oC= 27 = 273 = 300 K

To Find: r.m.s. speed = C₂ =?

Solution:

Now C₁ = 459.4 m/s, T₁ = 273 K, = T₂ = 27 ^oC= 27 = 273 = 300 K, C₂ = ?

Ans: The R.M.S. velocity of Oxygen molecules at 27 °C is 481.6 m/s

Example 11:

Compute the R.M.S. velocity of Oxygen molecules at 127 °C. Density of Oxygen at N.T.P. = 1.44 kg/m³.

Given: For nitrogen: Condition = N.T.P. T₁ = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², density of oxygen = r = 1.44 kg/m³, Temperature = T₂ = 127 ^oC= 127 = 273 = 400 K

To Find: r.m.s. speed = C₂ =?

Solution:

Now C₁ = 459.4 m/s, T₁ = 273 K, = T₂ = 127 ^oC= 127 = 273 = 400 K, C₂ =?

Ans: The R.M.S. velocity of Oxygen molecules at 127 °C is 556.1 m/s.

Example 12:

Calculate the R.M.S. velocity of Oxygen molecules at 225 °C. The density of oxygen at NTP is 1.42 kg/m⁵ and 1 atmosphere = 1.013 x 10⁵ N/m².

Given: For nitrogen: Condition = N.T.P. T₁ = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², density of oxygen = ρ = 1.44 kg/m³, Temperature = T₂ = 225 ^oC= 225 + 273 = 498 K

To Find: r.m.s. speed = C₂ =?

Solution:

Now C₁ = 459.4 m/s, T₁ = 273 K, = T₂ = 127 ^oC= 127 = 273 = 400 K, C₂ = ?

Ans: The R.M.S. velocity of Oxygen molecules at 127 °C is 624.8 m/s

Example 13: 

The density of a gas is 0.178 kg/m³ at N.T.P. Find the R.M.S. velocity of gas molecules. By what factor will the velocity of molecules increase at 200 °C?

Given: For nitrogen: Condition = N.T.P. T₁ = 273 K, P = 76 cm of Hg = 0.76 x 13600 x 9.8 = 1.013 x 10⁵ N/m², density of oxygen = ρ = 0.178 kg/m³, Temperature = T₂ = 200 ^oC= 200 + 273 = 473 K

To Find: r.m.s. speed = C₂/ C₁=?

Solution:

Now T₁ = 273 K, = T₂ = 200 ^oC= 200 + 273 = 473 K

Ans: The r.m.s. velocity of the gas molecule at NTP is 1.306 km/s. The r.m.s. velocity will increase by a factor of 1.316 at 200 °C.

Example 14: 

R M.S. velocity of oxygen molecules at 27 °C is 500 m/s. Calculate the R.M.S. and mean square velocities of oxygen molecules at 127 °C.

Given: C1 = 500 m/s at temperature T₁ = 27 ^oC = 27 + 273 = 300 K, Required speed at temperature = T₂ = 127 ^oC = 127 + 273 = 400 K,

To Find: C₂ =? and (C₂)²

Solution:

(C₂)² = (577.4)² = 3.33 x 10⁵ m²/s²

Ans: r.m.s. velocity is 577.4 m/s, and mean square velocity is 3.33 x 10⁵ m²/s²

Example 15:

Taking the R.M.S. velocity of Hydrogen molecules at N.T.P. as 1.84 km/s, calculate the R.M.S. velocity of Oxygen molecules at N.T.P. Molecular weights of Oxygen and Hydrogen are 32 and 2 respectively.

Given: r.m.s. velocity of hydrogen = C_H = 1.84 km/s, molecular mass M_O = 32, M_H = 2, temperature T_H = T_O = 273 K, pressure P_H = P_O = 1.013 x 10⁵ N/m².

To Find: r.m.s. velocity of oxygen molecule = C_O =?

Solution:

Ans: The R.M.S. velocity of Oxygen molecules at N.T.P. is 0.46 km/s

Example 16:

R.M.S. speed of Oxygen molecules is 493 m/s at a certain temperature. Calculate the R.M.S. speed of helium molecules at the same temperature. Molecular weights of Oxygen and Helium are 32 and 4 respectively.

Solution:

Given: r.m.s. velocity of oxygen = C_O = 493 m/s, molecular mass M_O = 32, M_He = 4,
temperature T_He = T_O, Pressure P_He = P_O.

To Find: r.m.s. velocity of helium molecule = C_He=?

Ans: The r.m.s. speed of helium molecule is 1394.2 m/s

Example 17:

R.M.S. speed of Oxygen molecules at N.T.P. is 459.3 m/s. Find the R.M.S. speed of Nitrogen molecules at 340 K. Molecular weights of Oxygen and Nitrogen are respectively 32 and 28.

Given: r.m.s. speed of oxygen = C_O1 = 459.3 m/s, T_O1 = 273 K, T_O2 = 340 K  = T_N , M_O = 32, M_N = 28

To Find: C_N =?

Solution:

Ans: r.m.s speed of nitrogen at 340 K is 548 m/s

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