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 2×105 Pa; D and C lie on the isobar at 1×105 Pa; A and D are at 300 K, B and C are at 500 K. A→B is isobaric (2×105 Pa, 300 K→500 K), B→C is isothermal (500 K), C→D is isobaric (1×105 Pa, 500 K→300 K), D→A is isothermal (300 K). n=2,γ=1.67.]
Assuming the gas to be ideal the work done on the gas in taking it from A to B is
(A)200 R
(B)300 R
(C)400 R✓ Correct
(D)500 R
Answer: (C)
Solution
The step from A to B takes place at constant pressure, where the work is simply W=nRΔT. With two moles and the temperature rising from 300 K to 500 K, this is 2R×(500−300)=400R.
JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option
