Question

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 328 K to 1312 K. If v1 is the average speed of the gas molecules before the absorption of heat and v2 their average speed after the absorption of heat, what is the ratio v2/v1?

Answer 1

Answer:

$\dfrac{v_2}{v_1}=2$

Explanation:

$k_b$ = Boltzmann constant = $1.38\times 10^{-23}\ J/K$

T = Temperature

m = Mass of the molecules

The RMS velocity of molecules in a gas is given by

$v_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

It can be seen that RMS velocty is proportional to temperature

$v_{rms}=\sqrt{T}$

$\dfrac{v_2}{v_1}=\sqrt{\dfrac{T_2}{T_1}}\\\Rightarrow \dfrac{v_2}{v_1}=\sqrt{\dfrac{1312}{328}}\\\Rightarrow \dfrac{v_2}{v_1}=2$

The ratio of the velocities is $\dfrac{v_2}{v_1}=2$