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**

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