Manipulate partial derivatives to obtain desired physical expression

[Roo2]

Homework Statement

Show that the expression A,

T(dP/dT)|V - P

is equal to expression B,

T^2 * [d(P/T)/dT]|V

Also, show that expression C,

-[d(P/T)/d(1/T)]|V

is also equal to expression B

Homework Equations

A: temperature * (dPresure/dTemperature at constant volume) - Pressue

B: Temperature^2 * d(Pressure/Temperature)/dTemperature at constant volume

C: - d(Pressure/Temperature)/d(1/Temperature) at constant volume

The Attempt at a Solution

Uhh... P/T = nR/V and 1/T = nR/PV... but I'm not sure how to keep the volumes and pressures constant in my solution. I feel like the second question should be especially simple but I just don't see the operations needed to do it. Could anyone help?

 

[cepheid]

The starting point is the ideal gas law, expressed in terms of P as a function of T:

P(T) = (nRT)/V

Now, if V = const., then P is just proportional to T, (i.e. P = const. * T), with the constant of proportionality given by: nR/V

Take expression A. Let's just focus on the partial derivative:

\frac{\partial P(T)}{\partial T}\bigg|_V = \frac{\partial}{\partial T}\left( \frac{nR}{V}T\right) \bigg|_V = \frac{nR}{V}

where the last step is just because the effect of differentiating w.r.t. T is to get rid of the factor of T and keep the constant out front.

Now, if you multiply this partial derivative by T, you just get back what you started with. If you then subtract P, what is THAT equal to?

 

[Roo2]

Mmk... So using that, I get the following relationship.

T(dP/dT)|V - P =

T(d[nRT/V]/dT] - P =

T(nR/V) - nRT/V = 0

Also,

T^2 d(P/T)/dT | V = T^2 (d[nR/V]/dT) = 0

Is this correct?

The starting point is the ideal gas law, expressed in terms of P as a function of T:

P(T) = (nRT)/V

Now, if V = const., then P is just proportional to T, (i.e. P = const. * T), with the constant of proportionality given by: nR/V

Take expression A. Let's just focus on the partial derivative:

\frac{\partial P(T)}{\partial T}\bigg|_V = \frac{\partial}{\partial T}\left( \frac{nR}{V}T\right) \bigg|_V = \frac{nR}{V}

where the last step is just because the effect of differentiating w.r.t. T is to get rid of the factor of T and keep the constant out front.

Now, if you multiply this partial derivative by T, you just get back what you started with. If you then subtract P, what is THAT equal to?

 

[cepheid]

Mmk... So using that, I get the following relationship.

T(dP/dT)|V - P =

T(d[nRT/V]/dT] - P =

T(nR/V) - nRT/V = 0

Also,

T^2 d(P/T)/dT | V = T^2 (d[nR/V]/dT) = 0

Is this correct?

Yes, I think that that is correct.

 

[Roo2]

Thanks!