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!

 

[HalfLostAndSearching]

The solutions manual may have made a mistake in their calculation for the heat absorbed on the path AB. Your calculation (11/2NKT1) seems to be correct, as it takes into account the change in internal energy and the work done by the gas. The solution manual's calculation (6NKT1) only takes into account the work done by the gas, but neglects the change in internal energy. This may be a typo or error in their solution. It is important to always double check your calculations and assumptions when solving thermodynamics problems.