Question

Two moles of an ideal monatomic gas occupies a volume v

Answer 1

Hey dear,
Given question is incomplete. Complete one is attached as image.

◆ Answer-
T2 = -84 °C
∆U = -2.7 kJ

◆ Explanation-
# Given-
T1 = 27 °C
V1 = V
V2 = 2V
n = 2
γ = 5/3.

# Solution-
We know the relation-
P ∝ V^-γ
P1 / P2 = V2^γ / V1^γ

But P = nRT/V, hence
(nRT1/V1) / (nRT2/V2) = V2^γ / V1^γ
T1/T2 = V2^(γ-1) / V1^(γ-1)
T2 = T1.(V2/V1)^(γ-1)
T2 = 300 × (2V/V)^(5/3-1)
T2 = 300 × 2^(2/3)
T2 = 189 K = -84 °C

Change in internal energy is -
∆U = (f/2)nR∆T
∆U = 3/2 × 2 × 8.314 × (189-300)
∆U = -2700 J = -2.7 kJ

Hope this helps you...