Find the change in entropy for an ideal gas undergoing a reversible process

[mcas]

Homework Statement

An ideal gad had a temperature ##T_1## and volume ##V_1##. As a result of a reversible process, these quantities changed to ##T_2## and ##V_2##. Find the change in entropy.

Relevant Equations

##pV=nRT##
##\delta Q = TdS##
##dU = \delta Q + \delta W##
##U = \frac{3}{2}kT##

We know that
$$dU=\delta Q + \delta W$$
$$dU = TdS - pdV$$
So from this:
$$dS = \frac{1}{T}dU + \frac{1}{T}pdV \ (*)$$
For an ideal gas:
$$dU = \frac{3}{2}nkdT$$
Plugging that into (*) and also from ##p=\frac{nRT}{V}## we get:
$$S = \frac{3}{2}nk \int^{T_2}_{T_1} \frac{1}{T}dT + R\int^{V_2}_{V_1} \frac{1}{V}dV$$

And so on...

Is this the correct approach to solve this problem? I'm not really sure because I'm still new to thermodynamics.

 

[TSny]

Your approach looks good. A couple of things, though.

##U = \frac{3}{2}kT##

This equation is for a monatomic ideal gas and it's missing a factor of ##N## (the number of molecules). But the question does not specify that the gas is monatomic. So, you'll need a more general expression for ##U## (usually expressed in terms of the number of moles ##n##, the molar heat capacity at constant volume, ##C_V##, and ##T##).

$$S = \frac{3}{2}nk \int^{T_2}_{T_1} \frac{1}{T}dT + R\int^{V_2}_{V_1} \frac{1}{V}dV$$

The first term should be corrected according to the remarks above. The second term is missing a factor. Can you spot it?

 

[mcas]

So, you'll need a more general expression for ##U## (usually expressed in terms of the number of moles ##n##, the molar heat capacity at constant volume, ##C_V##, and ##T##).

Ok, thank you. I think I know which one

The first term should be corrected according to the remarks above. The second term is missing a factor. Can you spot it?

I missed ##n##, right?

Thank you, this means very much! Now I have the motivation to do more problems

 

[TSny]

I missed ##n##, right?

Yes.