Change in entropy, quasistatic, isothermal expansion

[Fuffuf]

Homework Statement

I am to show that ΔS=Q/T for the isothermal expansion of a monoatomic ideal gas, when the expansion is so slow that the gas is always in equilibrium.

Homework Equations

1. law: ΔU=Q+W (We mustn't use dQ and dW - our teacher hates that :( ).
Ideal gas law: PV=NkT
We need the equation: ΔS=Nk*ln(V_final/V_initial)
And that quasistatic expansion work is W=-PΔV

The Attempt at a Solution

-I think I am to start with: ΔU=Q+W⇔Q=ΔU-W, where ΔU=0 since its isothermal.
-I know that it is quasistatic expansion work, so W = -PΔV, so Q = -(-PΔV) = PΔV
I think I want to get something from the ideal gas law in here: P=(NkT)/V, so

Q=(NkTΔV)/V

But then I kind of get stuck there...

Hope someone can help. I thinks it is really easy, but I kind wrap my head around it.

 

[RUber]

Looking at what you are supposed to show, take Q/T and compare that to what you have for $\Delta S$.
So it seems like the question is: does $\frac {\Delta V}{V} = \ln \frac{V_{final}}{V_{initial}}$?

 

[Fuffuf]

Yeah - something like that?

 

[RUber]

What do you know about the function for V and/or $\Delta$V if you are given an initial and final state?
Also, in the equations you provided for the ideal gas law, you change between R and k, are these different parameters?

 

[Fuffuf]

I don't know anything but what I have written unfortunately. No, sorry - that's my mistake. It should have been k all along.

 

[Chestermiller]

You really had the right idea in your initial post. Nice job. Now, just express the heat in differential form:
$$dQ=PdV=\frac{NkT}{V}dV$$
Then integrate between the initial and final volumes.

Chet

 

[Fuffuf]

Great - thanks a lot :)