GATE 2017 ME (ME1) – Question 43

Thermodynamics · Zeroth, first and second laws, work, heat and energy changes · 2 marks · Numerical answer

One kg of ideal gas has R = 400 J/(kg K), cᵥ = 1000 J/(kg K), and starts at 1 bar and 300 K in a sealed rigid cylinder. During an adiabatic process a stirrer does 100 kJ of work on it. Find its entropy increase in J/K.

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Correct answer: 285 to 289

Explanation

The vessel is rigid and adiabatic, so no boundary work or heat crosses it. Stirrer work raises internal energy: $mc_v(T_2-T_1)=100000$ J, giving $T_2=400$ K. At constant volume, $\Delta S=mc_v\ln(T_2/T_1)=1000\ln(400/300)=287.682$ J/K. Adiabatic stirring is irreversible and therefore is not isentropic.

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