Thermodynamic problem and formula homework help

[dUDEonAfORUM]

An ideal gas at initial state has temperature 300 K has been compressed under constant pressure of 30 Pa from volume 3 cubic meters to 1.8 cubic meters. In the process 75 J of heat was lost.
A. Find the change in internal energy
B. Find the final temperature

I tried two methods in A and they give out different answers
Method I
formula: ΔU = Q- W = Q - PΔV
answer -39 J

Method II
formula: ΔU = 3/2 nRΔT = 3/2 PΔV
since this is an isobaric process PΔV=nRΔT
But the answer from this is -54 Jwhich method is wrong? If Method II is wrong, but then when can I use ΔU=3/2 PΔV
I think Method II is wrong as P in PΔV=nRΔT refers to the pressure of the ideal gas, but the pressure given is the external pressure acting on the gas...is this true?

As for B, I don't know whether to use PΔV=nRΔT or ΔU = 3/2 nRΔT

 

[ramzerimar]

An ideal gas at initial state has temperature 300 K has been compressed under constant pressure of 30 Pa from volume 3 cubic meters to 1.8 cubic meters. In the process 75 J of heat was lost.
A. Find the change in internal energy
B. Find the final temperature

I tried two methods in A and they give out different answers
Method I
formula: ΔU = Q- W = Q - PΔV
answer -39 J

Method II
formula: ΔU = 3/2 nRΔT = 3/2 PΔV
since this is an isobaric process PΔV=nRΔT
But the answer from this is -54 Jwhich method is wrong? If Method II is wrong, but then when can I use ΔU=3/2 PΔV
I think Method II is wrong as P in PΔV=nRΔT refers to the pressure of the ideal gas, but the pressure given is the external pressure acting on the gas...is this true?

As for B, I don't know whether to use PΔV=nRΔT or ΔU = 3/2 nRΔT

If the question didn't specified if the gas is monoatomic, diatomic or so, you probably should stay safe and use method I.

EDIT: Also, I don't think you are accounting for the 75 J of energy lost due to heat transfer during the process when you use the second method.

 

[Chestermiller]

Method 2 is wrong for the very reason you gave. In an irreversible compression like this, you can't use the ideal gas law because (1) the gas pressure isn't even uniform within the cylinder during the deformation and (2) viscous stresses in the gas (due to the rapid deformation) contribute to the force per unit area that the gas exerts on the piston face.