(12)Chemistry  Science (Dr. Bob222)
posted by Ana .
Calculate delta E for a process in which 88.0 g of nitrous oxide gas (N2O) is cooled from 165 ˚C to 65 ˚C at a constant pressure of 5.00 atm. The molecular weight of nitrous oxide is 44.02 g/mol, and its heat capacity at constant pressure, Cp, is 38.70 J/molK.
A)  7740 J
B) + 1040 J
C)  6077 J
D) + 6077J
E)  1040 J

I believe, and I can be wrong, that you have to assume that VOLUME and Pressure remains constant.
E=work pressure*change volume+heat
Work (W)=pressure*change in volume
and
Heat (q)=ncdT
If there is no change in volume, then work=0 and the equation becomes the following:
E=q=ncdT
Where
n=moles of N2O
c=38.70 J/molK.
and
dT= change in temperature=TfTi
Solve for moles:
n=88.0 g of N2O/44.013 g
n=1.999 moles
Convert from Celsius to Kelvin
273.15+165=438.15K
and
273.15 + 65=338K
Tf=338.15K
and
Ti= 438.15K
dT=338.15K438.15K=100K
Solve for q:
q=(1.999 moles)(38.70J/mol*K)(100K)
q=7736 J
Best answer choice is A. 
Thank you Devron for this clarification.

I have solved it this way too and I got answer A: 7740 J, but the correct answer is actually C :(

...not sure how to derive it

The problem here is in your assumptions. You KNOW the volume will change if the pressure is constant AND the temperature decreases from 165 to 65. Volume is proportional to T; therefore, volume decreases with decreasing T.
dE = q + w
You've made the assumption that w is zero and you've calculated q which is 7740 J (and I obtained 7740 and not 7736 but I used n = 2 also). Then use PV = nRT to calculate V at 165 and V at 65. I obtained dV = 3.28L and that x 5.00 atm gives about 1640 L*atm and that converted to J (x 101.325) = about 1662 J. Then dE = q + w
dE = 7740J + 1662J = 6078 J
Note a shortcut here. Calculate pdV directly as follows:
pdV = nRdT
pdV = 2*0.08205*100*101.325 = 1663 J, then dE = 7740J + 1663 J = 6077 J and that is much neater. You don't make those rounding errors in reading V1 and V2 and subtracting. 
I initially did it without assuming that volume wasn't constant and I calculated what Dr. Bob222 calculated, but I assumed that it was about 15J, so I just assumed that it was incorrect without checking a source to see if I needed to use a conversion. That is why I said I believe!!!!. Luckily that Dr. Bob222 always comes along and double checks answers given by others.

This problem is quite confusing. Thank you for your contribution and elucidation!