A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure $P_{1} = 10^{5} P a$ and volume $V_{i} = 10^{- 3} m^{3}$ changes to final state at $P_{f} = \frac{1}{32} \times 10^{5} p a & V_{f} = 8 \times 10^{- 3} m^{3}$ in an adiabatic quasi-static process, such that $P^{3} V^{5} =$ constant. Consider another thermodynamic process that brings the system form the same initial state to the same final state in two steps: an isobaric expansion at $P_{i}$ followed by an isochoric (is volumetric) process at volume $V_{f}$ . The amount of heat supplied to the system in the two-step process is approximately

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112 J

294 J

813 J

588 J

answer is C.

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Detailed Solution

For adiabatic process

$P^{3} V^{5} = co n s n t$

$P V^{5 / 3} = co n s n t$

$r = 5 / 3$

For process AC

$Δ Q_{1} = n C_{p} Δ T = \frac{n 5}{2} R Δ T$

$= \frac{5}{2} P Δ V$

$Δ Q_{1} = 17.5 \times 10^{2} J = 1750 J$

For process CD

$Δ Q_{2} = n C_{v} Δ T = n (\frac{3}{2} R) = \frac{3}{2} V Δ P$

$Δ Q_{2} = 3 / 2 \times 8 \times 10^{- 3} \times (- 1 + \frac{1}{32}) \times 10^{5}$

$Δ Q_{2} = - 11.625 \times 10^{2}$

$Δ Q_{n e t} = 1750 - 1162 = 588 J$

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