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 s zero. The heat absorbed by the gas is 50 J. The increase in internal energy from a to b is
`Delta U = Delta Q =50J.`
As the internal energy is 1550 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 increase in internal energy from b to c is
`Delta U =-Delta W=40J.`
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
`Delta U =1500 J - 1590 J =-90J`
The heat given to the system is `Delta Q =-70J`
Using `Delta Q =Delta U + Delta W,`
`Delta W_(ca) = Delta Q - Delta U =-70 J + 90J =20J.`