A monatomic ideal gas at an initial temperature of 390 K is compressed adiabatically from an initial volume of 120 L to a final volume of 40.0 L. What is the final temperature of the gas?
A) 124 K
B) 811 K
C) 610 K
D) 775 K
E) 820 K

Question

Quizplus · Accepted Answer

Since the process is adiabatic, we have

P1V1^γ = P2V2^γ,

where γ = C_P/C_V is the ratio of specific heats for a monatomic ideal gas.

For the given initial and final volumes, we have

V2/V1 = 1/3,

so we can express the pressure ratio as

P2/P1 = (V1/V2)^γ = (3)^γ.

For an adiabatic process, we also have

P1V1^γ/T1 = P2V2^γ/T2,

so we can use the pressure ratio to eliminate P1, and solve for T2:

T2 = T1 * (V1/V2)^(γ-1)

= 390 K * (1/3)^(2/5)

≈ 811 K.

Therefore, the answer is B.

A Monatomic Ideal Gas at an Initial Temperature of 390