Question

Assume ideal gas behaviour for all the gases considered and neglect vibrational degrees of freedom. Separate equimolar samples of $Ne,O_2,C{O}_2$ and $S{O}_2$ were subjected to a two process as mentioned. Initially all are at same state of temperature and pressure.
$\text{Step}I\to$ All undergo reversible adiabatic expansion to attain same final volume, which is double the original volume thereby causing the decreases in their temperature.
$\text{Step}II\to$ After step $I$ all are given appropriate amount of heat isochorically to restore the original temperature.
Mark the correct option (s):

• A. Due to step $I$ only, the decrease in temperature will be maximum for $Ne$
• B. During step $II$, heat given will be minimum for $S{O}_2$
• C. There will be no change in internal energy for any of the gas after both the steps of process are completed
• D. The $P-V$ graph of $O_2$ and $C{O}_2$ will be same

Answer 1

Answer: (A), (C), (D)

The $P-V$ graph of $O_2$ and $C{O}_2$ will be same

(a) ${\gamma }_{S{O}_2}<{\gamma }_{C{O}_2}={\gamma }_{O_2}<{\gamma }_{Ne}$
We know $P^{1-\gamma }{T}^{\gamma }=const.$
As $\gamma$ is maximum for $Ne$, so from step $I$ the decrease in temperature will be maximum for $Ne$.


(b) For an isochoric process work done = 0
$\therefore \Delta E=Q$
More is the degree of freedom more is the internal energy and more will be the heat given.
$S{O}_2$ has maximum degree of freedom. So, heat given will be maximum for $S{O}_2$.

(c) There will be no change in the internal energy for any of the gas after both the processes are completed as internal energy is a state function and its temperature dependent only.
First step is adiabatic $(q=0)$ so $\Delta U_1=w_1$
Second step is isochoric $(w=0)$
So, $\Delta U_2=q_2$
$\because$ initial and final temp. are same
$\therefore \Delta U_{total}=\Delta U_1+\Delta U_2=0$
Or $w_1+q_2=0$
(d) As the $\gamma$ value of $O_2$ and $C{O}_2$ are same $(i.e.1.40)$, the $P-V$ graph for $O_2$ and $C{O}_2$ will be same.