Question

Two moles of a monoatomic ideal gas ($C_V=1.5R$) initially at 400 K in an isolated cylinder of volume 1.0 L with a freely movable piston is allowed to expand suddenly against a constant pressure of 1.0 atm till the final volume reaches to 10 L. Which of the following conclusions regarding the above changes is/are true?
(Given $R=0.08Latm{K}^{-1}mo{l}^{-1}$)

• A. Final temperature of the gas is 362.5 K.
• B. If the same process was carried out to the same final volume, but under reversible conditions, final temperature would have been less than 362.5 K.
• C. In the above process, the initial and final temperatures and volumes are related as
  $\left[\frac{T_1}{T_2}\right]=[\frac{V_2}{V_1}{]}^{\gamma -1}$
• D. Entropy change of the system $\left(\Delta S_{sys}\right)$ is zero.

Answer 1

Answer: (A), (B)

The correct options are
A Final temperature of the gas is 362.5 K.
B If the same process was carried out to the same final volume, but under reversible conditions, final temperature would have been less than 362.5 K.

We have the relation,
$\Delta U=q+w$
Since the process is adiabatic q=0.
$n{C}_V\Delta T=-P\Delta V\Rightarrow \frac{3}{2}R\times 2\times (T_2-400)=-1\times 9\Rightarrow 3\times 0.08\times (400-T_2)=9\Rightarrow T_2=362.5K$

Since reversible work done is the maximum work done, the temperature change during the adiabatic expansion will be more, i.e, more cooling will be observed. Thus the final temperature would have been less than 362.5 K in reversible conditions.

$\left[\frac{T_1}{T_2}\right]=[\frac{V_2}{V_1}{]}^{\gamma -1}$ this equation ONLY works for reversible adiabatic processes.

For an irreversible expansion, the decrease in temperature will be lower as the work done is lower. Hence, decrease in entropy due to a fall in temperature will be lower in case of an irreversible expansion. Thus net entropy will increase.