Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram

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

Options

A. `1076R`
B. `1904R`
C. Zero
D. `276R`

Answer

D

Explanation

`W_(ABCDA) = W_(AB) + W_(DC) + W_(CD) + W_(DC)`
=`muR(500-300) + 2.303muR(500)"log"(2)/(2) + muR(300-500) + 2.303muR(300)"log"(1)/(2) = 276R`
Work done on the gas = `-276R`