Argon in a pistoncylinder assembly is compressed from state

Argon in a piston-cylinder assembly is compressed from state 1, where T1 = 300K, V1 = 1 m3 to state 2 where T2 = 200K. If the change in specific entropy is -0.27 kJ/kg-K, determine the final volume in m3. Assume ideal gas model with k = 1.67.

Solution

>> As, dS = Entropy Change = Cp*ln(T2/T1) - R*ln[(V1*T2)/(V2*T1)]

Now, Here, R = 8.314/40 = 0.20785 [ 40 is Molar Mass of Argon ]

and, Cp = [k/(k-1)]*R = 1.67*0.20785/0.67 = 0.5181 KJ/Kg

>> As, it is given that, dS = - 0.27 KJ/Kg-K

and, T1 = 300 K

V1 = 1 m³

T2 = 200 K

V2 = ?

>> Puting all in above equation,

=> - 0.27 = 0.5181*ln(200/300) - 0.20785*ln[(1*200)/(V2*300)]

=> - 0.05993 = -0.20785*ln[2/(3*V2)]

Solving this,

=> V2 = 0.50 m³