Question

An ideal gas is at pressure P and temperature T in a box, which is kept in vacuum within a large container. The wall of the box is punctured. What happens as the gas occupies all the entire container?
A. Its temperature falls
B. Its temperature rises
C. Its temperature remains the same
D. Unpredictable

Answer 1

The work done during the free expansion of an ideal gas is always zero when the process is irreversible.
The internal energy for an ideal gas is dependent only on the temperature.

Complete step by step answer:
Given that, an ideal gas is at pressure P and temperature T in a box, which is kept in vacuum within a large container.
We need to find out the effect on the temperature of the system when the wall of the box is punctured and the gas occupies the entire container.
The free expansion of a gas takes place when the gas is subjected to expansion against zero pressure.
Since, in the present case, the ideal gas is expanding in vacuum, it indicates that the work or the expansion of the ideal gas is taking place against a pressure of zero.
Since in free expansion, there is no external pressure applied to the system therefore, ${{\text{P}}_{{\text{ext}}}} = 0$ .
Thus, the work done by the gas in an infinitesimal expansion ‘dw’ will be:
${\text{dw = - }}{{\text{P}}_{{\text{ext}}}}{\text{dV = 0}}$
Here, dV denotes the change in volume.
Thus, the total work ‘w’ done by the ideal gas in this free expansion is given by:
${\text{w = }}\int {{\text{dw}}} \\\ \Rightarrow {\text{w = }} - \int {{{\text{P}}_{{\text{ext}}}}{\text{dV}}} \\\ \Rightarrow {\text{w = }}0 \\\$
Thus, no work is done by the gas.
In this case, there is no absorption or evolution of heat. Therefore, the heat ‘q’ is also equal to zero.
Now, from the first law of thermodynamics, the change in internal energy is:
$\Delta {\text{U}} = {\text{q + w}} \\\ \Rightarrow \Delta {\text{U}} = 0 + 0 \\\ \Rightarrow \Delta {\text{U}} = 0 \\\$
Thus, there is no change in internal energy.
Since for an ideal gas, internal energy is a function of temperature, therefore, constant internal energy will mean constant temperature.

Thus, the temperature remains the same and so C is correct.

Note:
For reversible isothermal change, the work done can be expressed as:
${\text{w = - nRTln}}\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{V}}_{\text{1}}}}}$
Here, n is the number of moles of the gas, R is gas constant and ${{\text{V}}_{\text{1}}}$ and ${{\text{V}}_2}$ denotes the initial and final volumes respectively.
In case of irreversible adiabatic free expansion, the change in enthalpy which depends on the sum of the internal energy change and temperature will also become zero.