Helium of mass `m = 1.7 g` is expanded adiabatically `n = 3.0` times and then compressed isobarically down to the initial volume. Find the entropy increment of the gas in this process.

Explanation

If the initial temperature is `T_0` and volume is `V_0` then in adiabatic expansion.
`TV^(gamma - 1) = T_0 V_0^(gamma - 1)`
so, `T = T_0 n^(1 - gamma) = T_1` where `n = (V_1)/(V_0)`
`V_1` being the volume at the end of the adiabatic process. There is no entropy change in this process. Next the gas is compressed isobarically and the net entropy change is
`Delta S = ((m)/(M) C_p) 1n (T_f)/(T_1)`
But `(V_1)/(T_1) = (V_0)/(T_f)`, or `T_f = T_1 (V_0)/(V_1) = T_0 n^- gamma`
So `Delta S = ((m)/(M) C_p) 1n (1)/(n) = -(m)/(M) C_p 1n n = -(m)/(M) (R gamma)/(gamma - 1) 1n n = - 9.7 J//K`.