How Is Temperature Related to Pressure in Adiabatic Expansion of an Ideal Gas?

[nicksauce]

Homework Statement

Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation:
$$\frac{dT}{dP} = \frac{2T}{5P}$$

Homework Equations

Ideal gas law

The Attempt at a Solution

$$PV = nkT$$

$$T = \frac{PV}{nk}$$

$$\frac{dT}{dP} = \frac{1}{nk}(V + P\frac{dV}{dP})$$

So what is dV/dP?

$$PV^\gamma = C$$

$$V = C^{\frac{1}{\gamma}} P^{\frac{-1}{\gamma}}$$

$$\frac{dV}{dP} = C ^ {\frac{1}{\gamma}} \frac{-1}{\gamma}P^{\frac{-1}{\gamma} - 1}$$

$$\frac{dV}{dP} = \frac{-1}{\gamma} VP^{-1}$$
so

$$\frac{dT}{dP} = \frac{1}{nk}[V - \frac{V}{\gamma}]$$

But this gives dT/dP ~= 1/3 T/P (using gamma~=3/2), instead of 2/5 T/P. Can anyone see where I went wrong?

 

[nicksauce]

Err. never mind, gamma should be 5/3, not 3/2, in which case this does give the correct answer.

 

[Moetasim]

Dear student,

I believe you have made a small mistake in your calculation. In the step where you substitute for dV/dP, you have written P^(-1) instead of P^(-1/gamma). This leads to the incorrect result of 1/3 T/P. If you correct this mistake, you should get the desired result of 2/5 T/P. Keep up the good work!