Entropy of an Irreversible Process: Consider a piston containing one mole ideal gas $($argon$),$ the

gas in the cylinder is irreversibly compressed by an external pressure. The initial temperature and

molar volume are $($T$1,$ V$1)$ and the final T and V are $($T$2,$ V$2).$ CV and R are known constants.

a$.$ Derive an expression for $$S of the gas for this irreversible process by manipulating the

equation dU $=$ TdS $$ PdV$.$ Note that T$1,$ V$1,$ T$2,$ V$2,$ CV and R are known terms.

Some students suggest another way of calculating the $$S of the above process. They consider that

the gas in the piston undergoes two processes and reaches the same state at the end:

Process $1$: heat the gas reversibly at constant volume from $($T$1,$ V$1)$ to $($T$2,$ V$1)$; then

Process $2$: compress the gas reversibly and isothermally from $($T$2,$ V$1)$ to $($T$2,$ V$2).$

b$.$ Derive $$S following these students$$ method. You may need to conduct energy and entropy

balances for the two reversible processes.

c$.$ Compare your answer in part b with the one in part a$,$ and state whether you get the same

answer. Do you think the second method works, and why?