Question

One mole of ideal gas initially at 300 K is expanded from an initial pressure of 10 atm to a final pressure of 1 atm. Calculate ΔU, q, w, ΔH, and the final temperature T2 for this expansion carried out according to each of the following paths. The heat capacity of an ideal gas is cV=3R/2.

1. A reversible adiabatic expansion.

Answer 1

ANSWER
ΔU = q + w =====First law of thermodynamics

in adiabatic process q = 0

therefore ΔU = w

ΔU = C_v ΔT

ΔT = Tf - Ti

Final temperature can be found fy following equation

(T1/T2) = (P2/P1)^1-ϒ/ϒ

ϒ = Cp / Cv

Cp = Cv + R = 3R/2 + R = 5/2R

ϒ = (5/2R) /  (3R/2) = 1.66

(T1/T2)= (P2/P1)^{(1-1.66) / 1.66}

(300 /T2)= (1/10)^-0.397

(300 /T2)= (10)^0.397

(300 /T2)= 2.5

T2 = 300 / 2.5 = 120K

ΔT = Tf - Ti = 120 - 300 = -180K

ΔU = C_v ΔT

ΔU = 3/2 R X 180 =  3/2 X 8.314 X (-180) = - 2244.78J

w = ΔU = - 2244.78J

ΔU = Cp ΔT = 5/2 X 8.314 X (-180) = - 3741.3J