Thermodynamics - PV Diagram of an Ideal Gas

[Maybe_Memorie]

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.

Q_AB = ΔU + ΔW = ΔU + PΔV, U is internal energy, U = U(T) = C_VT

For a monatomic ideal gas, C_V = 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 Q_AB = 6NKT₁ in the proceeding problems as well so it's a little confusing.

Thanks for any help.

 

[Chestermiller]

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

Chet

 

[Andrew Mason]

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.

Q_AB = ΔU + ΔW = ΔU + PΔV, U is internal energy, U = U(T) = C_VT

For a monatomic ideal gas, C_V = 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 Q_AB = 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

 

[Maybe_Memorie]

Thank you!