Question

An ideal monoatomic 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 $x$. Ignoring the friction between the piston and the cylinder, the correct statement(s) is (are)

• 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)

The correct options are
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 $\frac{7}{3}{P}_1{V}_1$.

We will assume that the gas on the side of container having spring is connected to atmosphere. Hence, its pressure remains constant at $P_1$.
In final position of piston, force balance gives
$Kx+P_{1}A=P_{2}A$
where A = cross-sectional area of the piston.
So, $Kx=\left(P_2-P_1\right)A\left(i\right)$
Also,
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
If $V_2=2V_1,T_2=3T_1,$then $P_2=\frac{3}{2}{P}_1$
Also, $V_2-V_1=Ax\left(ii\right)$
$\left(i\right)$ and $\left(ii\right)$ give $Kx=\left(P_2-P_1\right)\frac{\left(}{V_2-V_1}\therefore \frac{1}{2}K{x}^2=\frac{1}{2}\left(P_2-P_1\right)\left(V_2-V_1\right)\left(iii\right)=\frac{1}{2}\left(\frac{P_1}{2}\right)\left(V_1\right)=\frac{P_1{V}_1}{4}$
Hence (a) is correct.
$\Delta U=\frac{f}{2}\left(P_2{V}_2-P_1{V}_1\right)=\frac{3}{2}\left(\frac{3}{2}\times 2P_1{V}_1-P_1{V}_1\right]=3P_1{V}_1$
Hence, (b) is correct as well.
Net work done on piston by all forces is zero. So,
$W_{gas}-\frac{1}{2}K{x}^2-P_1\left(V_2-V_1\right)=0\therefore W_{gas}=\frac{P_1{V}_1}{3}+2P_1{V}_1{W}_{gas}=\frac{7}{3}{P}_1{V}_1$
So, (c) also turns out be correct.

$Q=\Delta U+W\Delta U=\frac{3}{2}\left(P_2{V}_2-P_1{V}_1\right)=\frac{3}{2}\left(\frac{4}{3}{P}_1.3V_1-P_1{V}_1\right)\Rightarrow \Delta U=\frac{9}{2}{P}_1{V}_1\therefore Q=\frac{9}{2}{P}_1{V}_1+\frac{7}{3}{P}_1{V}_1\Rightarrow Q=\frac{41}{6}{P}_1{V}_1$
Hence (d) is not correct.