Internal energy won't add up to 0 in cyclic process

  • #1
giraffe714
19
2
Homework Statement
A 1.00-mol sample of an ideal monatomic gas is taken through the cycle shown in Figure P22.54. The process A -> B is a reversible isothermal expansion.
(e) Calculate the changes in internal energy for each process.
Relevant Equations
ΔU_isothermal = 0, ΔU_isobaric = nC_P*ΔT, ΔU_isochoric = nC_V*ΔT, ΔU = Q + W (maybe?), PV = nRT
My problem isn't exactly with calculating the actual changes in internal energy, I'll put those values below. My problem is that I can't get the values to add up to 0, and I don't understand why since for cyclic processes, by definition, ΔU must equal 0.

$$ΔU_{AB} = ΔU_{isothermal} = 0$$
$$ΔU_{BC} = ΔU_{isobaric} = nC_PΔT_{BC} = n\frac{R}{1 - 1/γ}(T_C - T_B) = 1\frac{8.314}{1 - 1/(5/3)}(122 - 609) = -10122 \text{ J}$$
$$ΔU_{CA} = ΔU_{isochoric} = nC_VΔT_{CA} = n\frac{R}{γ - 1}(T_A - T_C) = 1\frac{8.314}{5/3 - 1}(609 - 122) = 6073\text{ J}$$

Which doesn't add up to 0 like it should for a cyclic process. What am I missing here? I feel like something with the "reversible" isothermal process but I'm not sure? And in case you're wondering where I got the values 122 and 609 from for the temperatures, I calculated them using PV = nRT:

##T_A = \frac{P_A*V_A}{nR} = \frac{5*101325*10*0.001}{1*8.314} = 609\text{ K}## (here I'm converting atm to Pa and L to m^3)
Now since A -> B is isothermal ##T_B = T_A = 609\text{ K}##
##T_C = \frac{1*101325*10*0.001}{1*8.314} = 122\text{ K}##
 

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  • #2
Okay, so I figured it out, I was using the wrong equation for ΔU_isobaric, it's still ##nC_VΔT##, I got that formula confused with the one for heat transfer in an isobaric process, ##Q = nC_PΔT##.
 
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FAQ: Internal energy won't add up to 0 in cyclic process

Why doesn't internal energy add up to zero in a cyclic process?

In a cyclic process, the internal energy change over one complete cycle is indeed zero. The confusion often arises from not accounting for all forms of energy transfer, such as work done and heat exchanged, which must balance each other out to result in no net change in internal energy.

How can internal energy be zero if work is done in a cyclic process?

In a cyclic process, any work done by the system must be exactly balanced by the heat added to the system. This ensures that the internal energy, which is a state function, returns to its original value at the end of the cycle, making the net change zero.

What role does the first law of thermodynamics play in cyclic processes?

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In a cyclic process, this means that the total heat added must equal the total work done, resulting in no net change in internal energy.

Can internal energy be non-zero at any point in a cyclic process?

Yes, the internal energy can vary at different points within the cycle. However, the key point is that the internal energy returns to its initial value at the end of the cycle, resulting in a net change of zero over the entire process.

What might cause confusion about internal energy in cyclic processes?

Confusion often arises from a misunderstanding of how heat and work interact in a cyclic process. It's crucial to remember that while heat and work can vary throughout the cycle, their net effect over one complete cycle must balance out to result in no net change in internal energy.

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