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

Assuming the gas to be ideal the work done on the gas in taking it form A to B is :

Options

A. `-690R`
B. `+690R`
C. `-414R`
D. `+414R`

Answer

D

Explanation

`W_(DA)` = Work done in isothermal process = `2.303 muRT log(P_(1))/(P_(2))`
=`2.303 xx 2R xx 300 log(1xx10^(5))/(2xx10^(5))`
=`2.303 xx 2R xx 300 xx(-0.3010) = -414R`
Work done on the gas = `-(-414R) = 414R`