When work W is done on an ideal gas of N diatomic molecules in thermal isolation the temperature increases by:
A)W/2Nk
B)W/3Nk
C)2W/3Nk
D)2W/5Nk
E)W/Nk

Question

Quizplus · Accepted Answer

For a diatomic gas, the degrees of freedom in the absence of vibrational modes (which is a common assumption at room temperature) is 5 (3 translational and 2 rotational). Using the equipartition theorem, each degree of freedom contributes $ \frac{1}{2}kT $ to the energy, so the total energy change for a temperature change is $ \Delta E = \frac{5}{2}Nk\Delta T $. When work W is done on the gas in thermal isolation, $ W = \Delta E $, thus $ W = \frac{5}{2}Nk\Delta T $, solving for $ \Delta T $ gives $ \Delta T = \frac{2W}{5Nk} $.

When Work W Is Done on an Ideal Gas of N