Question

An ideal diatomic gas at 80 K is slowly compressed adiabatically to one-third its original volume. What is its final temperature?

a) 80 K
b) 160 K
c) 240 K
d) 320 K

Answer 1

The final temperature of an ideal diatomic gas after adiabatically compressing to one-third of its volume is found using the adiabatic process equation, considering a heat capacity ratio for diatomic gases of 7/5 and initial temperature of 80 K.

Step-by-step explanation:

The question asks for the final temperature of an ideal diatomic gas which is adiabatically and slowly compressed to one-third of its original volume. Using the adiabatic process equation T1 * V1^(y-1) = T2 * V2^(y-1), where y (gamma) is the heat capacity ratio (Cp/Cv) for diatomic gases, which is 7/5. For an adiabatic compression of an ideal diatomic gas initially at 80 K and then compressed to one-third of its original volume, we need to solve for the final temperature T2.

Let's denote the initial temperature as T1 (80 K) and the initial and final volumes as V1 and V2 (V2 = V1/3). The equation then becomes:

80 K * (V1)^(7/5 - 1) = T2 * (V1/3)^(7/5 - 1)

Solving for T2, we find the final temperature of the ideal diatomic gas after the adiabatic compression. The temperature increases as the volume decreases due to work being done on the gas, increasing its internal energy.