[thekid_frankie]

In this problem I am given
T_i= 273 K
T_f= 373 K
C_p/(J K^-1)= 20.17 + 0.4001(T/K)

And am asked to calculate q, W, ΔU, and ΔH at
a)  constant pressure
b)  constant volume
for a perfect gas.

I was able to calculate ΔH as follows
[tex]
{\Delta H} = {C_p \Delta T} = \int^{373K}_{273K}(a+b(T))dT = [aT + \frac{1}{2}bT^2]^{373K}_{273K} = 20.17(373-273)+\frac{1}{2}(0.4001)(373^2-273^2)= 14.94kJ
[/tex]

Since the change in enthalpy is equal to the energy supplied as heat at constant pressure
ΔH = q_p = 14.94kJ

Now we have enthalpy and heat, work and internal energy are left.  For work to be done, we need to have a changing volume with some external pressure.  The issue here is we aren't given a volume, pressure, or # moles.  Am I free to assume that since the volume is free to change ΔU = 0 and q = -W?  This would yield
ΔH = 14.94kJ
q = 14.94kJ
W = -14.94kJ
ΔU = 0kJ

And if this is correct, with constant volume, W = O and that would give an increase of internal energy of 14.94kJ?  Seems too easy to be able to make assumptions like this.
Thanks for any help/corrections!

[Enthalpy]

Heat (internal, enthalpy, capacity...) is proportional to the amount of matter. In the expression for Cp, J/K is incomplete: it lacks the unit of matter amount. Already "20" instead of 1000 or 10^-23 tells what unit it is - a unit often taken implicitly in thermodynamics, alas. Probably elsewhere in the text.

A given amount of a perfect gas can have A times more pressure, but then its pressure is A times smaller, and when you sum things like PdV or VdP the result doesn't depend on A, so you don't need P nor V, only the amount of matter. The sum depends just on the amount of matter and the variation of... of... no, colour not, age neither. I depends on its...

Then, you know relationships between Cp and Cv for one amount unit of perfect gas, or between H and E for one amount unit of perfect gas, and you also know how Cp, Cv, H and E relate to heat and work, for the two special cases of constant pressure and constant volume. dU<>0 in both cases.