Problem 15.10 Part A Consider the following two-step process. Heat is allowed to
ID: 1793315 • Letter: P
Question
Problem 15.10 Part A 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 P1-2.5 atm to P2 1.7 atm. Then the gas expands at constant pressure, from a volume of V1 = 5.7 L to ½-9.2 L Calculate the total work done by the gas in the process Express your answer to two significant figures and include the appropriate units. where the temperature reaches its original value See the figure (Figure 1) Figure 1 of 1 W-Value Units Submit My Answers Give Up Pi Part B Calculate the change in internal energy of the gas in the process P2 Express your answer with the appropriate units AUValueUnits 0 V V2 Submit My Answers Give UpExplanation / 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)
= - 1.7 * 101325Pa (0.0092m³ - 0.0057m³)
= -602.9 Pam³
= -602.9 J
So the total work in the two-step process is:
W = W + W = 0J + (-602.9)J = 602.9J
(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
(c)
Change of internal energy equals work done On the gas plus heat transferred to the gas:
U = W + Q
=>
Q = U - W = 0J - (-602.9J) = 602.9J
That means 602.9J of heat are flowing into the gas during the two-step process