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← JEE Advanced 2016 · Paper 2
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JEE Advanced 2016 · Paper 2, Physics question 3

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi=105P_i=10^5 Pa and volume Vi=10−3 m3V_i=10^{-3}\ \mathrm m^3 changes to a final state at Pf=(1/32)×105P_f=(1/32)\times10^5 Pa and Vf=8×10−3 m3V_f=8\times10^{-3}\ \mathrm m^3 in an adiabatic quasi-static process, such that P3V5=P^3V^5= constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at PiP_i followed by an isochoric (isovolumetric) process at volume VfV_f. The amount of heat supplied to the system in the two-step process is approximately

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Answer and worked solution

Official answer: C

The adiabatic path. P3V5P^3V^5 = constant means γ=53\gamma=\frac53.

Change in internal energy. PiVi=100P_iV_i=100 J and PfVf=10532×8×10−3=25P_fV_f=\frac{10^5}{32}\times8\times10^{-3}=25 J, so
ΔU=PfVf−PiViγ−1=25−1002/3=−112.5 J.\Delta U=\frac{P_fV_f-P_iV_i}{\gamma-1}=\frac{25-100}{2/3}=-112.5\text{ J}.
ΔU\Delta U depends only on the end states.

Two-step path. Work is done only in the isobaric step:
W=Pi(Vf−Vi)=105×7×10−3=700 J.W=P_i(V_f-V_i)=10^5\times7\times10^{-3}=700\text{ J}.

Heat. Q=ΔU+W=587.5≈588Q=\Delta U+W=587.5\approx588 J.

(C)

Topics:First law and internal energyThermodynamic processesWork done by a gas
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