Calculating Changes in Enthalpy and Entropy for Isobaric Gas Expansion

[Trenthan]

Homework Statement

Gas has a heat capacity of (C_v = (3/2)nR), intial temp of T_o, pressure P_o and volume V_o.

Process a to b is a quasistatic isobaric expansion to twice the intial volume

Find
a) work done,
b) heat transfer,
c) change in internal energy,
d) change in enthalpy
e) change in entropy
In terms of n(moles), R, T₀

Homework Equations

W = -P(delta)V
(delta)U = Q + W = (f/2)nR(delta)T
PV = nrT
H = U + PV

The Attempt at a Solution

Now this is what I've come up with but I am unsure about enthalpy and entropy (this mostly), they seem ok but look wrong. I don't have any answer or similar questions to check it againt first and only isobaric cycle. Looking on the net I've found mostly constant U, while V changes which isn't helpful.

Attached is my attemp if any1 can shine any light on the last two points or anything else

Thanks Trent

 

[diazona]

I thought the formula was
$$\Delta S = \int_{T_i}^{T_f} \frac{C_V}{T}\mathrm{d}T$$
but anyway, you seem to have the right idea with the calculations. :-)

 

[Trenthan]

I thought the formula was
$$\Delta S = \int_{T_i}^{T_f} \frac{C_V}{T}\mathrm{d}T$$
but anyway, you seem to have the right idea with the calculations. :-)

Im basing this $$\Delta S = \int_{T_i}^{T_f} \frac{C_P}{T}\mathrm{d}T$$, from this

"When T is changing, it's usually more convenient to write the relation in terms of the heat capacipty at constant volume: dS = (C_VdT)/T."
Now we don't have constant volume so I've discarded that

Over the page than it says,
"constant pressure processes in which the temperature changes, we can write Q = C_PdT, than integrate to obtain,

$$\Delta S_P= \int_{T_i}^{T_f} \frac{C_P}{T}\mathrm{d}T$$

This correct or i may have read misunderstood something, i agree it should be C_V rather than C_P, but from my understanding that's for constant volume??

If any1 can shine some light thanks heeps

Cheers Trent