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Thermodynamics  PV Diagram of an Ideal Gas
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.
PV = NKT for an ideal gas.
Q_{AB} = ΔU + ΔW = ΔU + PΔV, U is internal energy, U = U(T) = C_{V}T
For a monatomic ideal gas, C_{V} = 3/2
The change in internal energy is obviously just 3/2NK(4T1  T1) = 9/2NKT1
The work done is P_{1}(2V_{1}  V_{1}) = NKT_{1}
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)(2T_{1}) + 3/2P_{1}V_{1} = 6NKT_{1} which obviously isn't true so I'm questioning it's integrity. It uses the fact that Q_{AB} = 6NKT_{1} in the proceeding problems as well so it's a little confusing.
Thanks for any help.
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_{V}T
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 P_{1}(2V_{1}  V_{1}) = NKT_{1}
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)(2T_{1}) + 3/2P_{1}V_{1} = 6NKT_{1} which obviously isn't true so I'm questioning it's integrity. It uses the fact that Q_{AB} = 6NKT_{1} in the proceeding problems as well so it's a little confusing.
Thanks for any help.
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