One mole of a gas is subjected to two process AB and BC, one after the other as shown in the figure. BC is represented by `PV^(n) = constant`. We can conclude that (where T = temperature, W = work done by gas, V = volume and U = internal energy).

Options

A. `T_(A) = T_(B) = T_(C)`
B. `V_(A) lt T_(B), P_(B) lt P_(C)`
C. `W_(AB) lt W_(BC)`
D. `U_(A) lt U_(B)`

Answer

D

Explanation

By `PV = nRT`
`T_(B) gt T_(A)`
`W_(AB) = P_(0)(2V_(0)-V_(0))= P_(0)V_(0)`
`W_(BC) = RT1n(3/2) = 2P_(0)V_90[1n3-1n2]`
`=2P_(0)V_(0)(2.3030)(0.477-0.30)`
`=0.8142 P_(0)V_(0) lt W_(AB)`.