Question

An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston. 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 $\frac{7}{3}{P}_1{V}_1$

$P=P_1+\frac{Kx}{A}$
$P_2=\frac{3}{2}{P}_1$

Thus we have $x=\frac{V_1}{A}$
$\frac{3P_1}{2}=P_1+\frac{Kx}{A}$
$Kx=\frac{P_{1}A}{2}$
Energy of spring $=\frac{1}{2}K{x}^2=\frac{P_{1}A}{4}x=\frac{P_1{V}_1}{4}$
For the case of option B
$\Delta U=\frac{f}{2}(P_2{V}_2-P_2{V}_1)$
$=3P_1{V}_1$
For the case of option C
$P_f=\frac{4P_1}{3}$ $Kx=\frac{P_1}{3}A$
or $x=\frac{2V_1}{A}$

$W_{gas}=-(W_{P_{atm}}+W_{spring})$
$=(P_{1}Ax+\frac{1}{2}K{x}^2)$
$=(P_{1}A\frac{2V_1}{A}+\frac{1}{2}\frac{P_{1}A}{3}\frac{2V_1}{A})$
$=2P_1{V}_1+\frac{P_1{V}_1}{3}=\frac{7P_1{V}_1}{3}$
For the case of option D
$\Delta Q=W+\Delta U$
$=\frac{7P_1{V}_1}{3}+\frac{3}{2}(P_2{V}_2-P_1{V}_1)$
$=\frac{7P_1{V}_1}{3}+\frac{3}{2}(\frac{4}{3}{P}_1.3V_1-P_1{V}_1)$
$=\frac{7P_1{V}_1}{3}+\frac{9}{2}{P}_1{V}_1=\frac{41P_1{V}_1}{6}$