Consider the following two-step process. Heat is allowed to flow out of an ideal
ID: 1505471 • Letter: C
Question
Consider the following two-step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from P1 = 2.3 atm to P2 = 1.0 atm . Then the gas expands at constant pressure, from a volume of V1 = 6.3 L to V2 = 9.8 L , where the temperature reaches its original value.
Part A
Calculate the total work done by the gas in the process.
Express your answer to two significant figures and include the appropriate units.
Part B
Calculate the change in internal energy of the gas in the process.
Express your answer with the appropriate units.
Explanation / Answer
(a)
The total work done is the sum of the work done in each step.
Work done on the gas is given by the integral
W = - p dV from initial to final volume
Since the the volume does not change in first step, no work is done on the gas:
W = 0
In second step gas expands at constant pressure, so the work integral simplifies to:
W = - p dV from initial to final volume
= - p (V_final - V_initial)
= - 1.0 *(101325Pa) *(0.0098m³ - 0.0063m³)
= -354.64Pam³
= -354.64J
So the total work in the two-step process is:
W = W + W = 0J + (-354.64)J = -354.64J
(b)
Assuming ideal gas behavior, internal energy depends solely on temperature (and not on pressure and volume):
U = nCvT
So the change of internal energy is:
U = nCvT
Because we return to initial temperature after second step, the change of internal energy is zero
U = 0