JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Heat & Thermodynamics

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure

P_{i} = 1 0^{5} Pa

and volume

V_{i} = 1 0^{- 3} m^{3}

changes to a final state at

P_{f} = (1/32) \times 1 0^{5} Pa

and

V_{f} = 8 \times 1 0^{- 3} m^{3}

in an adiabatic quasi-static process, such that

P^{3} V^{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

P_{i}

followed by an isochoric (isovolumetric) process at volume

V_{f}

. The amount of heat supplied to the system in the two-step process is approximately

(A)112 J
(B)294 J
(C)588 J✓ Correct
(D)813 J

Answer: (C)

Solution

From

P^{3} V^{5} = constant

, we identify

γ = 5/3

(monoatomic gas).

For the two-step process:

Work done in isobaric expansion:

W = P_{i} Δ V = 1 0^{5} \times 7 \times 1 0^{- 3} = 700

J

Change in internal energy:

Δ U = \frac{3}{2} (P_{f} V_{f} - P_{i} V_{i}) = \frac{3}{2} (\frac{1 0 ^{5} \times 8 \times 1 0 ^{- 3}}{32} - 1 0^{5} \times 1 0^{- 3}) = - 112.5

J

From first law:

H = W + Δ U = 700 - 112.5 \approx 588

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option

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JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Heat & Thermodynamics

Solution

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