- #1
Aldnoahz
- 37
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This is a question on my review package, but I still cannot understand the solution and do not know why my solution is wrong.
The PV diagram is attached.
Problem: During change AB, 300 J of thermal energy is supplied to the gas. During change BC, 250 J of thermal energy is transferred. The area ABC on the PV diagram represents 120 J of energy.
Calculate the thermal energy transfer during the stage CA.
I am using equation Q = U+W
W>0 when gas expands and DOES work and W<0 when work is DONE ON gas
2. The attempt at a solution
I did this question in two ways, arriving at the same answer, but neither is correct.
Way 1:
Since CA is an isochoric change, W=0 and so Q=change in U
To find change in U, first consider change AB:
Q= deltaUAB + W, Q=300J, W=120J (from graph)
deltaUAB = 180J
Then consider change BC:
Q= deltaUBC + W, Q=250J, W= -(120+120) = -240 J (area under curve)
deltaUBC = 490J
deltaUCA = deltaUAB + deltaUBC = 670 J = Q
Way 2:
Total energy put in = 300J + 250 J = 550J
120 J is done ON the gas, Wnet = - 120J
Q= deltaU + W
deltaU = Q - W = 550 - (-120) = 670 JHowever, the answer is 550 - 120 = 430J without much explanation. I am confused...
Homework Statement
The PV diagram is attached.
Problem: During change AB, 300 J of thermal energy is supplied to the gas. During change BC, 250 J of thermal energy is transferred. The area ABC on the PV diagram represents 120 J of energy.
Calculate the thermal energy transfer during the stage CA.
I am using equation Q = U+W
W>0 when gas expands and DOES work and W<0 when work is DONE ON gas
2. The attempt at a solution
I did this question in two ways, arriving at the same answer, but neither is correct.
Way 1:
Since CA is an isochoric change, W=0 and so Q=change in U
To find change in U, first consider change AB:
Q= deltaUAB + W, Q=300J, W=120J (from graph)
deltaUAB = 180J
Then consider change BC:
Q= deltaUBC + W, Q=250J, W= -(120+120) = -240 J (area under curve)
deltaUBC = 490J
deltaUCA = deltaUAB + deltaUBC = 670 J = Q
Way 2:
Total energy put in = 300J + 250 J = 550J
120 J is done ON the gas, Wnet = - 120J
Q= deltaU + W
deltaU = Q - W = 550 - (-120) = 670 JHowever, the answer is 550 - 120 = 430J without much explanation. I am confused...