An ideal gas is taken round a cyclic thermodynamic process ABCA as shown if Fig. If the internal energy of the gas at point A is assumed zero while at B it is 50 J. The heat absorbed by the gas in the process BC is 90 J.
(a) What is the internal energy og the gas at point C ?
(b) How much heat energy is absorbed by the gas in the process AB?
(c ) Find the heat energy rejected or absorbed by the gas in the process CA.
(d) What is the net work done by the gas in the complete cycle ABCA ?

Explanation

Given that `U_(A) = 0, U_(B) = 50 J` and `Q_(BC) = 90 J`
Also `P_(A) = P_(B) = 10 Nm^(-2), P_(C) = 300 Nm^(-2), V_(A) = 1m^(3)`
and `V_(B) = V_(C ) = 3 m^(3)`
a. In process BC as volume of gas remains constant, work done by gas in this process is zero, thus
`W_(BC) = 0`
Heat abosrbed by the gas is `Q_(BC_ = 90 J`. From the first law of thermodynamics.
`(Delta U) B_(C ) = U_(C ) - U_(B) = Q_(BC) - W_(BC) = 90 J - 0 = 90 J`
`U_(C ) = (Delta U)_(BC) + U_(B) = 90 J + 50 J = 140 J`
b. In process AB, we have
`(Delta U)_(AB) = U_(B) - U_(A)`
`= 50 - 0 = 50 J`
work done is given as
`W_(AB)` = area under AB in P - V diagram
= area of rectangle ABED
`= AB xx AD = (3 m^(3) = 1 m^(3)) xx 10 Nm^(-2)`
`= 20 J`
Thus heat abosrbed by the system is
`Q_(AB) = (Delta U)_(AB) + W_(AB) = 50 + 20 = 70 J`
c. For process CA
`(Delta U)_(CA) - U_(A) - U_(C ) = 0 - 140 = - 140 J`
Work done is given as
`W_(CA) = ` area ACED
= area of triangle ACB + area of rectangle ABED
`1//2 xx AB xx BC + AB xx AD`
`= 1//2 xx (3 - 1) m^(3) xx (30 - 10) Nm^(-2) + 20`
`= 20 + 20 = 40 J`
In this process, the volume decreases, the work is done on the gas, Hence, the work done is negaitve,
Thus,
`W_(CA) = - 40 J`
Thus heat rejected by the gas is
`Q_(CA) = (Delta U)_(CA) + W_(CA) = - 140 - 40 = - 180 J`
Net work done in the complete cyclic process ABCA is
`W = "area of triangle" ABC = (1)/(2) xx 2 xx 20 = 20 J`
As the cycle is anticlockwise, net work is done on the gas.