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 ${\mathrm{T}}_1$ , pressure ${\mathrm{P}}_1$ and volume ${\mathrm{V}}_1$ and the spring is in its relaxed state. The gas is then heated very slowly to temperature ${\mathrm{T}}_2$ , pressure ${\mathrm{P}}_2$ and volume ${\mathrm{V}}_2$ . During this process the piston moves out by a distance $\mathrm{x}$ . Ignoring the friction between the piston the cylinder, the correct statement(s) is (are)

• Correct answer
  A

  If ${\mathrm{V}}_2=2{\mathrm{V}}_1$ and ${\mathrm{T}}_2=3{\mathrm{T}}_1$ , then the energy stored in the spring is $\frac{1}{4}{\mathrm{P}}_1{\mathrm{V}}_1$
• Correct answer
  B

  If ${\mathrm{V}}_2=2{\mathrm{V}}_1$ and ${\mathrm{T}}_2=3{\mathrm{T}}_1$ , then the change in internal energy is $3{\mathrm{P}}_1{\mathrm{V}}_1$
• Correct answer
  C

  If ${\mathrm{V}}_2=3{\mathrm{V}}_1$ and ${\mathrm{T}}_2=4{\mathrm{T}}_1$ , then the work done by the gas is $\frac{7}{3}{\mathrm{P}}_1{\mathrm{V}}_1$
• D

  If ${\mathrm{V}}_2=3{\mathrm{V}}_1$ and ${\mathrm{T}}_2=4{\mathrm{T}}_1$ , then the heat supplied to the gas is $\frac{17}{6}{\mathrm{P}}_1{\mathrm{V}}_1$

Solution

(i) $\mathrm{P}={\mathrm{P}}_1+\frac{\mathrm{K}\mathrm{x}}{\mathrm{A}}$
${\mathrm{P}}_2=\frac{3}{2}{\mathrm{P}}_1\Rightarrow \mathrm{x}=\frac{{\mathrm{V}}_1}{\mathrm{A}}$
$\frac{3{\mathrm{P}}_1}{2}={\mathrm{P}}_1+\frac{\mathrm{K}\mathrm{x}}{\mathrm{A}}$
$\mathrm{K}\mathrm{x}=\frac{{\mathrm{P}}_1\mathrm{A}}{2}$
Energy of spring
$\frac{1}{2}\mathrm{K}{\mathrm{x}}^2=\frac{{\mathrm{P}}_1\mathrm{A}}{4}\mathrm{x}=\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{4}$
(ii) $∆U=\frac{f}{2}\left(P_2{V}_2-P_2{V}_1\right)$
$=3P_1{V}_1$
(iii) $P_f=\frac{4P_1}{3}$
$KX=\frac{P_1}{3}A$
$X=\frac{2V_1}{A}$
$W_{gas}=-\left(W_{P_{atm}}+W_{spring}\right)$
$=\left(P_{1}Ax+\frac{1}{2}Kx.x\right)$
$=+\left(P_{1}A.\frac{2V_1}{A}+\frac{1}{2}.\frac{P_{1}A}{3}.\frac{2V_1}{A}\right)$
$=2P_1{V}_1+\frac{P_1{V}_1}{3}=\frac{7P_1{V}_1}{3}$
(iv) $∆Q=W+∆U$
$=\frac{7P_1{V}_1}{3}+\frac{3}{2}\left(P_2{V}_2-P_1{V}_1\right)$
$=\frac{7P_1{V}_1}{3}+\frac{3}{2}\left(\frac{4}{3}{P}_1.3V_1-P_1{V}_1\right)$
$=\frac{7P_1{V}_1}{3}+\frac{9}{2}{P}_1{V}_1=\frac{41P_1{V}_1}{6}$

Hence, the correct answer is options: 'A', 'B', 'C'.