JEE Main · 2013 · 07 Apr 2013 · Offline · Heat & Thermodynamics

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The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is

(A) $(\frac{13}{2}) p_{0} v_{0}$ ✓ Correct
(B) $(\frac{11}{2}) p_{0} v_{0}$
(C) $4 p_{0} v_{0}$
(D) $p_{0} v_{0}$

Answer: (A)

Solution

Read the cycle off the

p

V

rectangle: the gas runs

D (v_{0}, p_{0}) \to A (v_{0}, 2 p_{0}) \to B (2 v_{0}, 2 p_{0}) \to C (2 v_{0}, p_{0}) \to D

. Heat is drawn from the source only on the legs where the temperature rises. On

D \to A

the volume is held at

v_{0}

while the pressure rises

p_{0} \to 2 p_{0}

, so

Q_{D A} = n C_{V} Δ T = \frac{3}{2} V Δ p = \frac{3}{2} p_{0} v_{0}

. On

A \to B

the pressure is held at

2 p_{0}

while the volume rises

v_{0} \to 2 v_{0}

, so

Q_{A B} = n C_{P} Δ T = \frac{5}{2} p Δ V = \frac{5}{2} (2 p_{0}) (v_{0}) = 5 p_{0} v_{0}

. The return legs

B \to C

and

C \to D

are cooling and compression, on which heat is rejected, not absorbed. The heat extracted from the source in one cycle is therefore

\frac{3}{2} p_{0} v_{0} + 5 p_{0} v_{0} = \frac{13}{2} p_{0} v_{0}

, using

C_{V} = \frac{3}{2} R

and

C_{P} = \frac{5}{2} R

for a monoatomic gas.

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option

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JEE Main · 2013 · 07 Apr 2013 · Offline · Heat & Thermodynamics

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