# Thermaldynamics - Adiabatic system W = delta(U)

1. Sep 24, 2011

### steve233

1. The problem statement, all variables and given/known data

After integrating the pressure formula of an adiabatic system, I have to show how this is equal to the change in energy. I know that my integral is correct (it was very straight forward), but I'm having trouble showing that it is equal to $\Delta$U.

2. Relevant equations

$\Delta$U = W + Q (Q = 0)
PV = NkT
$\Delta$U = 1/2 * N * k * f * $\Delta$T

Where:
k = 1.381 * 10-23
f = degrees of freedom
$\Delta$T = change in temperature

3. The attempt at a solution

W = -PiVi(Vf1-$\varphi$ - Vi1 - $\varphi$) / 1 - $\varphi$

Given that, I need to somehow get that to be
$\Delta$U = 1/2 * N * k * f * $\Delta$T

I managed to reduce W to:

-NkTi((Vf / Vi)1 - $\varphi$ - 1) / 1 - $\varphi$

But I'm stuck from there.
(Note: $\varphi$ = (f + 2) / f )
Any hints on what to do next would be very helpful.
Thanks

2. Sep 25, 2011

### rude man

Ideal gas:
pV = NkT
pV^(γ-1) = constant = C for adiabatic proc.
where γ = (f+1)/f

1. write expression for V(T) using both the above eq.
2. get dV = dV(N,k,γ,dT)
3. dW = p(V)dV(N,k,γ,dT). Note that c and V do not appear in this.