Thermodynamics - PV Diagram of an Ideal Gas

  • #1
Thermodynamics -- PV Diagram of an Ideal Gas

Homework Statement



Right, I've got a PV diagram attached to make this easier to explain. The substance is an ideal gas.

I need to calculate the heat absorbed along the different paths ACB, ADB and AB. The first two are easy. The red lines in the diagram are isotherms.

Homework Equations



PV = NKT for an ideal gas.

QAB = ΔU + ΔW = ΔU + PΔV, U is internal energy, U = U(T) = CVT

For a monatomic ideal gas, CV = 3/2

The Attempt at a Solution



The change in internal energy is obviously just 3/2NK(4T1 - T1) = 9/2NKT1

The work done is P1(2V1 - V1) = NKT1

Thus the heat absorbed on the path AB is 11/2NKT1.


My problem is the solution manual says it's 6NKT1.

It also says that ΔU + ΔW = 3/2(NK)(2T1) + 3/2P1V1 = 6NKT1 which obviously isn't true so I'm questioning it's integrity. It uses the fact that QAB = 6NKT1 in the proceeding problems as well so it's a little confusing.

Thanks for any help.
 

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Answers and Replies

  • #2
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The answer book is correct. For path AB, the average pressure is 3/2P1, and the volume change is V1.

Chet
 
  • #3
Andrew Mason
Science Advisor
Homework Helper
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Homework Statement



Right, I've got a PV diagram attached to make this easier to explain. The substance is an ideal gas.

I need to calculate the heat absorbed along the different paths ACB, ADB and AB. The first two are easy. The red lines in the diagram are isotherms.

Homework Equations



PV = NKT for an ideal gas.

QAB = ΔU + ΔW = ΔU + PΔV, U is internal energy, U = U(T) = CVT

For a monatomic ideal gas, CV = 3/2

The Attempt at a Solution



The change in internal energy is obviously just 3/2NK(4T1 - T1) = 9/2NKT1

The work done is P1(2V1 - V1) = NKT1

Thus the heat absorbed on the path AB is 11/2NKT1.


My problem is the solution manual says it's 6NKT1.

It also says that ΔU + ΔW = 3/2(NK)(2T1) + 3/2P1V1 = 6NKT1 which obviously isn't true so I'm questioning it's integrity. It uses the fact that QAB = 6NKT1 in the proceeding problems as well so it's a little confusing.

Thanks for any help.
Chestermiller is right. The area under the graph from A-B is (P1+P2)(V2-V1)/2 = 3P1V1/2.

However, if the book says ΔU + ΔW = 3/2(NK)(2T1) + 3/2P1V1 , that is not correct. ΔU = 3/2(NK)(3T1) as you have noted.

AM
 
  • #4
Thank you!
 

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