Question

An ideal gas can be expanded from an initial state to a certain volume through two different processes: (i) $P{V}^2$ = constant and (ii) $P=K{V}^2$, where $K$ is a positive constant. Then:

• A. Final temperature in (i) will be greater than in (ii).
• B. Final temperature in (ii) will be greater than in (i).
• C. Heat supplied in process (ii) is less than heat supplied in process (i).
• D. Heat supplied in process (ii) is greater than heat supplied in process (i).

Answer 1

Answer: (B), (D)

The correct options are
B Final temperature in (ii) will be greater than in (i).
D Heat supplied in process (ii) is greater than heat supplied in process (i).

Pictorial representation of the two processes is shown in the figure below.
Let the two processes be represented by $AC$ and $AB$.


Given, $V_C=V_B$ [final volumes same]
From the graph, we can deduce that
$P_C<P_B$
Using $PV=nRT$, we can say that $T_B>T_C$
$\Rightarrow T_B-T_A>T_C-T_A$

Also, $W_{AB}>W_{AC}$
[from area under the $P-V$ graph]

So, from the 1st law of thermodynamics,
$Q=\Delta U+W$
We can say that,
$Q_{AB}=\frac{f}{2}R\Delta T_{AB}+W_{AB}$
$Q_{AC}=\frac{f}{2}R\Delta T_{AC}+W_{AC}$
$\Rightarrow Q_{AB}>Q_{AC}$
$[\because (\Delta T{)}_{AB}>(\Delta T{)}_{AC}\text{and}{W}_{AB}>W_{AC}]$

Hence, options (B) and (D) are the correct answers.