JEE Main · 2009 · 12 Apr 2009 · Offline · Heat & Thermodynamics

Passage

Two moles of helium gas are taken over the cycle ABCDA, as shown in the P–T diagram. [Figure — cycle ABCDA in the P–T plane: A and B lie on the isobar at

2 \times 1 0^{5}

Pa; D and C lie on the isobar at

1 \times 1 0^{5}

Pa; A and D are at 300 K, B and C are at 500 K. A

\to

B is isobaric (

2 \times 1 0^{5}

Pa, 300 K

\to

500 K), B

\to

C is isothermal (500 K), C

\to

D is isobaric (

1 \times 1 0^{5}

Pa, 500 K

\to

300 K), D

\to

A is isothermal (300 K).

n = 2, γ = 1.67

The net work done on the gas in the cycle ABCDA is

(A)Zero
(B)276 R✓ Correct
(C)1076 R
(D)1904 R

Answer: (B)

Solution

Going once round the cycle, the work is the sum over the four legs. The isobaric legs contribute

W_{A B} = n R Δ T = 2 R (500 - 300) = 400 R

and

W_{C D} = 2 R (300 - 500) = - 400 R

, which cancel. The isothermal legs contribute

W_{B C} = n R T ln \frac{P _{i}}{P _{f}}

at 500 K and

W_{D A}

at 300 K, and combining them leaves

1000 R ln 2 - 600 R ln 2 = 400 R ln 2

. Numerically

400 R \times 0.693

gives about

276 R

for the net work over the cycle ABCDA.

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option

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JEE Main · 2009 · 12 Apr 2009 · Offline · Heat & Thermodynamics

Solution

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