Question

The r.m.s. speed of the molecules of a gas at a pressure $10^{5}$ Pa and temperature $0{^\circ}C$ is $0.5km\mspace{6mu}\sec^{- 1}.$ If the pressure is kept constant but temperature is raised to 819°C, the velocity will become

• A. 1.5 kms^–1
• B. 2 kms^–1
• C. 5 kms^–1
• D. 1 kms^–1

Answer 1

Answer: (D)

1 kms^–1

Answer 2

$\frac{(V_{rms})_{1}}{(v_{rms})_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$Ž $\frac{500}{(v_{rms})_{2}} = \sqrt{\frac{0 + 273}{819 + 273}} = \sqrt{\frac{273}{1092}}$

$(v_{rms})_{2} = 500\sqrt{\frac{1092}{273}} = 500\sqrt{4} = 1000\frac{m}{\sec\frac{km}{\sec}}$