V₁ = Al, ℓ = initial length of segment containing monoatomic gas.

${\text{P}}_{\text{2}} =\frac{{\mathrm{T}}_2}{{\mathrm{V}}_2}\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}$

(A)     ${\text{P}}_{\text{2}} =\frac{3{\mathrm{T}}_1}{2{\mathrm{V}}_1}\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{{\mathrm{T}}_1}=\frac{3}{2}{\mathrm{P}}_1$

P₁A + Kx  = P₂A

$\mathrm{kx}=({\mathrm{P}}_2-{\mathrm{P}}_1)\mathrm{A}=(\frac{3}{2}-1){\mathrm{P}}_1\mathrm{A}=\frac{{\mathrm{P}}_1\mathrm{A}}{2}$

as v₂ = 2v₁ so x = ℓ

$\text{so k}\mathrm{\ell }=\left(\frac{{\mathrm{P}}_1\mathrm{A}}{2}\right)$

energy of spring = $\frac{1}{2}{\mathrm{kx}}^2=\frac{1}{2}{\mathrm{k\ell }}^2=\frac{1}{2}\frac{{\mathrm{P}}_1\mathrm{A\ell }}{2}=\frac{{\mathrm{P}}_1{\mathrm{V}}_1}{4}$

increase in U, $\mathrm{\Delta U}=\frac{3}{2}$nR (2T₁) = 3nRT₁ = 3P₁V₁

Hence q = $∆$U + w = 3P₁V₁ +$\frac{{\mathrm{P}}_1{\mathrm{v}}_1}{4}=\frac{13{\mathrm{P}}_1{\mathrm{v}}_1}{4}$

(C)        For v2 = 3v₁ and T₂ = 4T₁

${\text{P}}_{\text{2}} =\frac{4}{3}{\text{ P}}_{\text{1}}$     $\Rightarrow \mathrm{kx}=(\frac{4}{3}-1){\mathrm{P}}_1\mathrm{A}=\frac{{\mathrm{P}}_1\mathrm{A}}{3}$            

work done by gas = P₁ (2ℓ)A +$\frac{1}{2}\frac{{\mathrm{P}}_1\mathrm{A}}{3}$ (2ℓ) = 2P₁v₁ + $\frac{{\mathrm{P}}_1{\mathrm{v}}_1}{3}=\frac{7}{3}={\mathrm{P}}_1{\mathrm{V}}_1$