General advice, well meant:
Always hang on the dimensions until done. Anything physical has a dimension (unless expressed as a ratio, and often even then!).
- You catch mistakes in expressions a lot earlier
- You catch unit conversion problems a lot earlier
- You catch things that cancel each other a lot earlier
- Accuracy improves almost automatically
- You need less time to re-calculate after discovering an error on the way
- You score more points if you nevertheless make a calculation error
And a few more things. In this case, you would have prevented ehilds pointing out that ΔH expression is wrong, because you would have seen it yourself already. On two counts:
You would have seen that Cv given is the
molar heat capacity, not the specific heat of your (huge) sample of Argon, which is unfortunately the Cv in ΔU = CvΔT.
You would also have questioned the second term, or rather the combination of that and the ...kj (please use kJ, not kj) -- and then corrected with a factor of 1000.And this good practice of keeping track of the units begins with summing up the given data:
T1=523.15 K
T2=823.15 K
M= 0.03995 kg
/mol
m=50 kg
R=8.31
J/(mol.K)
Cv=0.0125 kJ/(mol.K)
Your rendering of the original problem gives me the impression that this is only part of the whole exercise, but I can't be sure. It does break off somewhat strangely.
Reason: even in the 1.5 line of the problem statement there is too much already, so you have to select the useful stuff.
You either have learned already, or will learn soon (for instance in this exercise, if it is as long as I suspect), that H is useful in dealing with constant pressure cases, because its d/dT includes pV work. U, on the other hand, is very useful when no work is done, which is often the case in constant volume cases. There dU = dq + dw becomes dU = dq and with your ΔU=CvΔT equation you are fully in business -- provided you keep a keen eye on the units!
