Derivation of ideal gas heat capacity relationship

[jasonRF]

TL;DR

Do not understand a derivation in Physical Chemistry by McQuarrie and Simon.

The text derives C_p-C_v=nR for ideal gasses. They start with $$H = U + PV = U + nRT$$ for ideal gas. Since U is only a function of temperature for an ideal gas, the right-hand side is only a function of temperature so $$\frac{dH}{dT} = \frac{dU}{dT} + nR$$. Now the text does something I don't understand.

First they set $$\frac{dH}{dT} = \left( \frac{\partial H}{\partial T}\right)_p = C_p$$ Why can they assume constant pressure here? I feel like I am missing something fundamental.

Similarly, they set $$\frac{dU}{dT} = \left(\frac{\partial U}{\partial T} \right)_V = C_v$$. Again, I don't understand why they can assume constant volume.

Any help would be much appreciated.

Thanks!
Jason

 

[kuruman]

It's a matter of definition, not assumption.

$C_p$ denotes specific heat at constant pressure.
$C_v$ denotes specific heat at constant volume.

 

[jasonRF]

It's a matter of definition, not assumption.

$C_p$ denotes specific heat at constant pressure.
$C_v$ denotes specific heat at constant volume.

Thanks for the reply. I don't think I asked my question very clearly. I understand the definitions of $C_v$ and $C_p$. What I don't understand is why $$\frac{dH}{d T} = \left(\frac{\partial H}{\partial T}\right)_P$$ is true.

 

[Chestermiller]

For an ideal gas, H and U depend only on T.

 

[vanhees71]

You have
$$\mathrm{d} U = T \mathrm{d} S - p \mathrm{d} V,$$
and thus
$$C_{\text{V}}=T \left (\frac{\partial S}{\partial T} \right)_V=\left (\frac{\partial U}{\partial T} \right)_{V}.$$
Further from $H=U+pV$ you get
$$\mathrm{d} H = T \mathrm{d} S + V \mathrm{d} p,$$
and thus
$$C_{\text{p}}=T \left (\frac{\partial S}{\partial T} \right)_p=\left (\frac{\partial H}{\partial T} \right)_{p}.$$

 

[kuruman]

TL;DR Summary: Do not understand a derivation in Physical Chemistry by McQuarrie and Simon.

The text derives C_p-C_v=nR for ideal gasses. They start with $$H = U + PV = U + nRT$$ for ideal gas. Since U is only a function of temperature for an ideal gas, the right-hand side is only a function of temperature so $$\frac{dH}{dT} = \frac{dU}{dT} + nR$$. Now the text does something I don't understand.

First they set $$\frac{dH}{dT} = \left( \frac{\partial H}{\partial T}\right)_p = C_p$$ Why can they assume constant pressure here? I feel like I am missing something fundamental.

Similarly, they set $$\frac{dU}{dT} = \left(\frac{\partial U}{\partial T} \right)_V = C_v$$. Again, I don't understand why they can assume constant volume.

Any help would be much appreciated.

Thanks!
Jason

Perhaps a detailed derivation is in order.

Case I: Constant volume
Start with
$H=U+nRT$ (definition with ideal gas)
Then
$(dH)_V=(dU)_V+nR~dT$
The first law says
$(dU)_V=(dQ)_V$ because the work done by the gas is zero.
Then$$(dH)_V=(dQ)_V+nR~dT \implies \left(\frac{\partial H}{\partial T}\right)_V=\left(\frac{\partial Q}{\partial T}\right)_V+nR=C_V+nR.$$Case II: Constant pressure
Start with
$H=U+pV$ (definition)
Then
$(dH)_p=(dU)_V+d(pV)=(dU)_p+p~dV.$
The first law says
$(dU)_p=(dQ)_p-pdV$
Then$$(dH)_p=(dQ)_p \implies \left(\frac{\partial H}{\partial T}\right)_p=\left(\frac{\partial Q}{\partial T}\right)_p=C_p.$$As @Chestermiller already remarked, $H$ depends only on temperature, i.e. it doesn't matter whether the enthalpy is changing under constant volume or constant pressure. Therefore $$\left(\frac{\partial H}{\partial T}\right)_V=\left(\frac{\partial H}{\partial T}\right)_p$$ Hence, $$C_p=C_V+nR.$$

 

[jasonRF]

Thanks everyone. It makes more sense to me now. My brain can be slow sometimes!

Cheers!

Jason