Question

What is the entropy change when 1.20 g of propane (C3H8) at 0.100 atm pressure is compressed by a factor of 6 at a constant temperature of 20 ∘C? Assume that propane behaves as an ideal gas

Answer 1

Given, mass of propane = 1.20 g

moles(n) of propane will be = 1.20 g C₃H₈ x ( 1 mol C₃H₈ / 44 g C₃H₈) = 0.027 mol C₃H₈

P = 0.100 atm

T = 20⁰C = (20+273) K = 293 K

We also know that R = 8.314 J/mol.K

From ideal gas,

PV = nRT

=> PV/nT= R = constant

=> P_fV_f/nT = P_iV_i/nT

it is mentioned that P_f =6P_i ( i for initial and f for final)

Now,

P_fV_f =  P_iV_i

=> V_f /V_i = P_i / P_f

=> V_f /V_i =  P_i / 6P_i

=> V_f /V_i = 1/6

Entropy change is given by the relation as:

S = nRln (V_f /V_i)

=> S = 0.027 mol x 8.314 J/mol.K x ln (1/6)

=> S = 0.224478 x ln (0.1667) J/K

=> S = 0.224478 x -1.791559489

=> S = -0.402165691 J/K = - 0.402 J/K

Hence, the answer is  - 0.402 J/K .