C.V. Methane gas of constant mass m2 = m1 = m and reversible process.
(a) Process: 1Q2 = 0 => s2 = s1
thus isentropic process s = const and ideal gas gives relation
(b) Process: T = constant. For ideal gas then u2 = u1 and
Energy eq. gives 1W2 = 1Q2 and ∫ dQ/T = 1Q2/T
with the entropy change found
=> 1W2 = 1Q2 = mT(s2 - s1) = -mRT ln(P2/P1)
= -0.51835× 293.2 ln(800/100) = -316.0 kJ
(c) Process: Pvn = constant with n = 1.15 ;
The T-P relation is given