Adiabatic Expansion: Relating T & P with dT/dP

[w3390]

Homework Statement

Show that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation: dT/dP = (2/f+2)(T/P).

Homework Equations

PV=NkT
VT^(f/2) = constant
V^(gamma)*P = constant

The Attempt at a Solution

I started off with the formula for ideal gases, PV=NkT.

I rearranged to get T=(PV/Nk).

At this point I don't know where to go. I don't see any equations I can use to make substitutions and I'm not sure if I should take the derivative at this point or not?

 

[kuruman]

Use the ideal gas law to replace V in the second "relevant equation" that you posted.
Solve for T in terms of p.
Take the required derivative dT/dp.

 

[w3390]

Okay. I'm getting a little tripped up at the derivative part. I have at this point:

T = (c/Nk*P)^(2/f+2) , where c is a constant

Taking the derivative will bring down the 2/f+2, but that leaves me with (2/f+2)-1 as the exponent plus the derivative of the inside.

 

[kuruman]

A constant is a constant is a constant so you can write

$$T=CP^{\frac{2}{f+2}}$$

Then you say that

$$\frac{dT}{dP}=C\frac{2}{f+2}\:P^{\frac{2}{f+2}-1}$$

I am not sure what you mean by "the derivative of the inside." What do you get when you simplify the exponent? How is that related to the expression of T as a function of P?

 

[paranormal]

To solve for dT/dP, we need to use the ideal gas law and the relation between volume and temperature during adiabatic expansion.

Starting with the ideal gas law, we can rearrange it to get P = NkT/V.

Next, we can use the relation VT^(f/2) = constant, where f is the number of degrees of freedom of the gas molecules. For an ideal gas, f = 3, so we have VT^(3/2) = constant.

Since we are dealing with adiabatic expansion, we know that no heat is transferred, so the change in entropy is zero. This means that PV^(gamma) = constant, where gamma is the adiabatic index, which is equal to (f+2)/f.

Now, we can substitute for V in the equation PV^(gamma) = constant using our previous equations. This gives us P(NkT)^(gamma) = constant.

Rearranging, we get T^gamma = (constant)/P^(gamma-1).

Taking the natural log of both sides, we get gamma*ln(T) = ln(constant) - (gamma-1)ln(P).

Now, we can differentiate both sides with respect to P to get dT/dP = (gamma-1)/P * T^(gamma-2) * (-1/P).

Substituting for gamma = (f+2)/f and rearranging, we get dT/dP = (2/f+2) * (T/P).

Therefore, we have shown that when an ideal gas expands adiabatically, the temperature and pressure are related by the differential equation: dT/dP = (2/f+2)(T/P).