Question

A sample of neon gas is initially at 1.5 L, 300K, and 1.5 atm. It is simultaneously heated and expanded to 500K and 3.5 L. Calculate the entropy change for the gas using two different reversible paths.

Answer 1

P x V = n x R x T

1.5 atm x 1.5 L = n x 0.082 L.atm/K.mol x 300K

n = 2.25 atm.L / 24.6 L.atm/mol

n= 0.091 mol Ne

Cp = 5 x R / 2

Cp = 5 x 8.31 J/molK / 2

Cp = 20.775 J/molK

1.- Case A: If the process is an adiabatic transformation, then S=0; because there is no interchange of heat and these transformations are called isentropic.

2.- Case B: If the process is an isobaric process; then S = n x Cp x Ln(T₂/T₁)

S = 0.091mol x 20.775J/molK x Ln (500K/300K)

S = 0.9657 J/mol