Molar heat capacities: 3.0 moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at constant pressure of 6.9 cal/(mol ∙ K) starts at 300 K and is heated at constant pressure to 320 K, then cooled at constant volume to its original temperature. How much heat flows into the gas during this two-step process?

Question

Quizplus · Accepted Answer

First, we need to calculate the heat absorbed during heating the gas at constant pressure from 300 K to 320 K. We use the formula:

q = nCpΔT

Where, 
n = 3.0 moles (given) 
Cp = 6.9 cal/(mol∙K) (given) 
ΔT = (320 K - 300 K) = 20 K

Therefore,

q1 = (3.0 mol) x (6.9 cal/(mol∙K)) x (20 K) = 414 cal

Next, we need to calculate the heat released during cooling the gas at constant volume back to its initial temperature of 300 K. We use the formula:

q = nCvΔT

Where, 
n = 3.0 moles (given) 
Cv = 4.9 cal/(mol∙K) (given) 
ΔT = (320 K - 300 K) = -20 K (negative because the gas is being cooled)

Therefore,

q2 = (3.0 mol) x (4.9 cal/(mol∙K)) x (-20 K) = -294 cal

(Note that the heat released is negative because it is being released from the gas)

Finally, we can calculate the total heat absorbed by the gas during the two-step process as:

q = q1 + q2 = 414 cal - 294 cal = 120 cal

Therefore, the best choice is D, 120 cal.

Molar Heat Capacities: 3