Question

An ideal monatomic gas is confined in a horizontal cylinder by a spring - loaded piston (as shown in the figure). Initially the gas is at temperature $T_1$, pressure $p_1$ and volume $V_1$ and the spring is in its relaxed state. The gas is then heated very slowly to temperature $T_2$, pressure $p_2$ and volume $V_2$.
During this process the piston moves out by a distance $x$. Which of the following statements is /are correct?
(Ignore the friction between the piston and the cylinder).

• A. If $V_2=2V_1$ and $T_2=3T_1,$ then the energy stored in the spring is $\frac{1}{4}{p}_1{V}_1$.
• B. If $V_2=2V_1$ and $T_2=3T_1,$ then the change in internal energy is $3p_1{V}_1$
• C. If $V_2=3V_1$ and $T_2=4T_1,$ then the work done by the gas is $\frac{7}{3}{p}_1{V}_1$.
• D. If $V_2=3V_1$ and $T_2=4T_1,$ then the heat supplied to the gas is $\frac{17}{6}{p}_1{V}_1$.

Answer 1

Answer: (A), (B), (C)

If $V_2=3V_1$ and $T_2=4T_1,$ then the work done by the gas is $\frac{7}{3}{p}_1{V}_1$.

For $\left(a\right),\left(b\right)$
$V_2=2V_1$ and $T_2=3T_1$ $\Rightarrow p_2=\frac{3}{2}{p}_1$
$p=p_1+\frac{Kx}{A}$
$p_2=\frac{3}{2}{p}_1\Rightarrow x=\frac{V_1}{A}$
$\frac{3}{2}{p}_1=p_1+\frac{Kx}{A}$
$Kx=\frac{p_{1}A}{2}$
Energy stored in the spring $=\frac{1}{2}K{x}^2=\frac{1}{2}\left(Kx\right)x=\frac{1}{4}{p}_1{V}_1$.
Now, $\Delta U=\frac{f}{2}(p_2{V}_2-p_1{V}_1)=3p_1{V}_1$.

In option (c),(d):
If $V_2=3V_1$ and $T_2=4T_1$
$\Rightarrow p_2=\frac{4}{3}{p}_{1}Kx=\frac{p_1}{3}A,$
so $x=\frac{2V_1}{A}$
Now, $W_{gas}=-(W_{\text{atm}}+W_{\text{spring}})=p_{1}Ax+\frac{1}{2}K{x}^2$
$=|p_{1}A.\frac{2V_1}{A}+\frac{1}{2}.\frac{p_{1}A}{3}.\frac{2V_1}{A}|=\frac{7}{3}{p}_1{V}_1.$
Heat supplied to the system,
$\Delta Q=W+\Delta U=\frac{7}{3}{p}_1{V}_1+\frac{3}{2}(p_2{V}_2-p_1{V}_1)=\frac{41}{6}{p}_1{V}_1.$

Hence, the correct options are (a), (b) and (c)