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JEE Advanced 2023 Paper 2 · Physics · Work, Energy and Power · hard

A horizontal cylinder with a frictionless piston of area A contains a diatomic ideal gas (rigid rotor, Cv=5/2R) at pressure p0 and volume V0. The…

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A horizontal cylinder with a frictionless piston of area AA contains a diatomic ideal gas (rigid rotor, Cv=52RC_v=\tfrac52R) at pressure p0p_0 and volume V0V_0. The outside of the piston is exposed to atmospheric pressure p0p_0 and is also attached to a spring whose other end is fixed to a wall; initially the spring is at its natural length. The spring constant is k=p0A2/V0k=p_0A^2/V_0. The gas is heated slowly until its volume becomes 2V02V_0. The heat absorbed by the gas is Q=n p0V0Q=n\,p_0V_0. Find nn.
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Answer: 9

If the piston is displaced by xx, the volume is V=V0+AxV=V_0+Ax and force balance gives pA=p0A+kxpA=p_0A+kx, so p=p0+k(V−V0)A2=p0+p0(V−V0)V0=p0VV0p=p_0+\dfrac{k(V-V_0)}{A^2}=p_0+\dfrac{p_0(V-V_0)}{V_0}=p_0\dfrac{V}{V_0}. The process is p∝Vp\propto V, ending at V=2V0V=2V_0, p=2p0p=2p_0. Work done by the gas: W=∫p dV=p0V0⋅(2V0)2−V022=1.5 p0V0W=\int p\,dV=\dfrac{p_0}{V_0}\cdot\dfrac{(2V_0)^2-V_0^2}{2}=1.5\,p_0V_0. Change in internal energy: ΔU=52(pfVf−piVi)=52(4−1)p0V0=7.5 p0V0\Delta U=\tfrac52(p_fV_f-p_iV_i)=\tfrac52(4-1)p_0V_0=7.5\,p_0V_0. By the first law Q=ΔU+W=9 p0V0Q=\Delta U+W=9\,p_0V_0, so n=9n=9.
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