Question 803: Thermodynamics & KTG

Question

A fixed mass of gas at constant pressure occupies a volume ' V '. The gas undergoes a rise in temperature so that the r.m.s. velocity of the molecules is doubled. The new volume will be

Accepted Answer

For a fixed mass of gas at constant pressure, the volume occupied is 'V'. When the temperature rises, the root mean square (r.m.s.) velocity of the gas molecules doubles.
The r.m.s. velocity $\mathrm{V}_{	ext{rms}}$ is given by:
$ \mathrm{V}_{	ext{rms}} = \sqrt{\frac{3 \mathrm{KT}}{\mathrm{M}}} $
This implies:
$ \mathrm{V}_{	ext{rms}}^2 \propto \mathrm{T} $
Therefore, if the r.m.s. velocity of the molecules is doubled:
$ \mathrm{T}_2 = 4\mathrm{T} \quad 	ext{...(i)} $
At constant pressure, the volume of the gas is directly proportional to the temperature:
$ V \propto T $
Thus, we have the relationship:
$ \frac{V_1}{V_2} = \frac{T_1}{T_2} $
Substituting the known values from equation (i):
$ \frac{V}{V_2} = \frac{T}{4T}  $
Solving for $V_2$:
$ V_2 = 4V $

Answer

$\frac{V}{2}$

Answer

$\frac{\mathrm{V}}{\sqrt{2}}$

Answer

$2 \mathrm{~V}$

Answer

4 V

Thermodynamics & KTG

A fixed mass of gas at constant pressure occupies a volume ' V '. The gas undergoes a rise in temperature so that the r.m.s. velocity of the molecules is doubled. The new volume will be

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