(12)Chemistry - Science (Dr. Bob222)

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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

• (12)Chemistry - Science (Dr. Bob222) -

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=Tf-Ti

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=338.15K-438.15K=100K

Solve for q:

q=(1.999 moles)(38.70J/mol*K)(-100K)

q=-7736 J

• (12)Chemistry - Science (Dr. Bob222) -

Thank you Devron for this clarification.

• (12)Chemistry - Science (Dr. Bob222) -

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

• (12)Chemistry - Science (Dr. Bob222) -

...not sure how to derive it

• (12)Chemistry - Science (Dr. Bob222) -

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.

• (12)Chemistry - Science (Dr. Bob222) -

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.

• (12)Chemistry - Science (Dr. Bob222) -

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

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