Net heat in a thermodynamic cycle

In summary, the conversation involves a discussion of the first law of thermodynamics and the calculation of heat and work in an isothermal process. The process involves a gas at a constant temperature and the values of heat and work are determined by various equations. However, there is an inconsistency with one of the values, which is later discovered to be due to an incorrect pressure value. The conversation ends with the realization that this mistake could have been caused by a poorly constructed question in a national level entrance exam.
  • #1
Kaguro
221
57
Homework Statement
In the thermodynamic cycle shown, one mole of monoatomic ideal gas is taken through the cycle. AB is a reversible isothermal expansion at a temperature of 800K in which the volume of the gas is doubled. BC is an isobaric contraction to the original volume in which the temperature is reduced to 300K. CA is a constant volume process in which the pressure and temperature return to their initial values. Find the net amount of heat absorbed .
Relevant Equations
PV=RT
Q=C(T2-T1)
dU=dQ-dW
Cv=1.5R for monoatomic ideal gas
Cp = Cv+R = 2.5R
cycle.png


In AB, dU=0 (Isothermal)
Hence dQ=dW
W1= RT ln(2V/V) = RT ln(2) = 800R*ln(2)
Q1=W1 = 800R*ln(2)

Q2= Cp(300-800)=-500Cp = -500*2.5R

Q3 = Cv(800-300) = 500Cv = 500*1.5R

So Q= 800R*ln(2) -500*2.5R + 500*1.5R = 800R*ln(2) -500R
Hence net heat absorbed is 800R*ln(2) -500R

But.

Suppose I find work done instead:

W1 = 800R*ln(2)
W2 = P2(ΔV) = P2*(V-2V)=-P2*V
W3=0

Now, P1V1=RT
P2V2 = RT
So, P2=P1/2=0.5P1

So,W2=-0.5*P1*V1=-0.5*R*T1 = -400R

So, Net work done is 800*R*ln(2) - 400R
As this is a cycle, dU=0 over it. And net heat absorbed is net work done. Hence
Net heat absorbed is 800*R*ln(2) - 400R

There is inconsistency here. What did I do wrong?
 
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  • #2
The 300 K is inconsistent with the pressure being half the original pressure at point C. It should be 400 K.
 
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Likes Kaguro
  • #3
:oops: I didn't notice that..

It's very annoying to have national level entrance exams having wrong questions. :mad:

Thanks very much for pointing it out.
 

FAQ: Net heat in a thermodynamic cycle

What is net heat in a thermodynamic cycle?

Net heat in a thermodynamic cycle refers to the total amount of heat that is added to or removed from a system during a complete cycle. It includes both the heat added to the system and the heat rejected from the system.

How is net heat calculated in a thermodynamic cycle?

Net heat is calculated by taking the difference between the heat added to the system and the heat rejected from the system. This can be represented by the equation: Q_net = Q_in - Q_out, where Q_net is the net heat, Q_in is the heat added to the system, and Q_out is the heat rejected from the system.

What is the significance of net heat in a thermodynamic cycle?

The net heat in a thermodynamic cycle is important because it determines the efficiency of the cycle. The higher the net heat, the more work can be obtained from the system, resulting in a more efficient cycle.

How does net heat affect the performance of a thermodynamic cycle?

The net heat has a direct impact on the performance of a thermodynamic cycle. A higher net heat means more work can be obtained from the system, resulting in a more efficient cycle. On the other hand, a lower net heat will result in a less efficient cycle.

What factors can influence the net heat in a thermodynamic cycle?

The net heat in a thermodynamic cycle can be influenced by several factors, such as the type of working fluid used, the temperature and pressure of the system, and the design and efficiency of the cycle. Other external factors, such as environmental conditions, can also affect the net heat.

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