Internal Energy and Entropy?

Question

A rigid tank of small mass contains 55.0 g of argon, initially at 230°C and 100 kPa. The tank is placed into a reservoir at 0°C and is allowed to cool to thermal equilibrium. Calculate the following:

(a) The volume of the tank
**I found to be 57.59 L, which is correct. The rest is what I need help on...

(b) the change in internal energy of the argon
I know that the change in internal energy is equal to Q-W and thet W=Qhot-Qcold, but I'm not really sure what to plug in where to get this answer.

(c) the energy transferred by heat
Not even sure where to start with this...

(d) the change in entropy of the argon
I know that the change in entropy is equal to dQ/T and that
dQ = mcdT. I'm just really not sure what to use for what-- and what the change in temperature would be. Would the Argon cool to 0*C?

(e) the change in entropy of the constant-temperature bath

Answer 1

Dear Flossie,
You need a Physical Chemistry Book, and if you are happy with Partial Differentials then many of the equations are very easy.

a) pV = nRT you have about 1.5 mols of Argon.
b) Du = Cv.Dt. I assume you are given the Specific Heat Capacity of Argon at contant Volume in the Question(it is 2.98)
c) DQ = DU + pdV. As the volume does not change the second term is Zero
d) dS = Cv, dT/T. Integrate up to give: DS = Cv. {Ln(T2)-Ln(T1)}
e) DS = DQ/T = DU/T = DU/273

where D means delta or 'change of'
and d means differential or partial differential

Look this all up in a Book as I have written this partly from memory, but I hope this gives you a start.

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