Final answer:
In an adiabatic process where an ideal diatomic gas at 80 K is compressed to half its volume, the final temperature of the gas is calculated to be 160 K, which corresponds to option b) 160 K.
Step-by-step explanation:
The question is concerned with an adiabatic process involving an ideal diatomic gas. The process is reversible, and the gas is initially at 80 K and then compressed to half its volume without heat exchange with the environment. To calculate the final temperature of the gas, we can use the adiabatic equation for diatomic gases:
T1 * V1^(y-1) = T2 * V2^(y-1)
where T1 is the initial temperature, V1 is the initial volume, y is the heat capacity ratio (which is 7/5 for diatomic gases), T2 is the final temperature, and V2 is the final volume. Since the volume is halved, V2 = V1/2 and T1 is given as 80 K.
Plugging in the values, we get:
80 K * (V1)^(7/5-1) = T2 * (V1/2)^(7/5-1)
T2 = 80 K * 2^(7/5-1)
T2 = 80 K * 2^(2/5)
T2 = 80 K * (1.3195)
T2 = 160 K
Therefore, the final temperature of the gas after the adiabatic compression is 160 K. The mention correct option in the final part of our solution is option b) 160 K.