A sample of an ideal gas is taken through the cyclic process abca . It absorbs 50 J of heat during the part ab, no heat during bc and rejects 70 J of heat during ca. 40 J of work is done on the gas during the part bc.(a) find the internal energy of the gas at b and c if it is 1500 J at a. (b) calculate the work done by the gas during the part ca.

Explanation

(a) In the part ab the volume remains constant. Thus, the work done by the gas is zero. The heat absorbed by the gas is 50 J. the increase in internal energy from a to b is `DeltaU=DeltaQ=50 J`.
As the internal energy is 1500 J at a, it will be 1550 J at b. in the part bc, the work done by the gas is `DeltaW=-40J` and no heat is given to the system. the increse in internal energy from b to c is
`DeltaU=-DeltaW=40 J`. ,brgt As the internal energy is 1550 J at b, it will be 1590 J at c.
(b) The change in internal energy from c to a is
`DeltaU=1500 J-1590 J=-90J`.
The heat given to the system is `DeltaQ=70 J`.
Using `DeltaQ=DeltaU+DeltaW`,
`DeltaW=DeltaQ-DeltaU`
`=-70 J+90J=20 J`.