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
`Delta U = Delta Q = 50 J`
As the internal energy is `1500 J` at `a`, it will be `1500 J` at `b`. In the part `bc`, the work done by the gas is `Delta W = - 40 J` and no heat is given to the system. The increase in internal energy from b to c is `Delta U - Delta W = 40 J`
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 = - 90 J`
The heat given to the system is `Delta Q = - 70 J`.
Using `Delta Q = Delta U + Delta W`,
`Delta W ca = Delta Q - Delta U = - 70 J + 90 J = 20 J`