Ideal gas volume work expression (adiabatic)

[krootox217]

Homework Statement

I have the following task:

Homework Equations

The Attempt at a Solution

But I don't understand how to solve it. Can somebody help me?[/B]

 

[stockzahn]

1) I'd guess they want you to apply the first law of thermodynamics, but I'm not sure.
2) Here you just have to do what the statement says (with a substitution considering ideal gas).
3) Based on the result in 2) you are able to proof that.

You need the ideal gas equation and the connectedness between R, c_p and c_v

 

[vinicius0197]

First one: The differential form of the 1º law of thermodynamics is:
dU = dQ-dW
Because it's an adiabatic process, there is no heat exchange, so Q = 0.
dU = -dW
Internal energy is a function of state that depends solely on the quantity of gas - a number n of moles - and the absolute temperature. That can be written as:
dU = nC_vdT
As for work:
dW = Fdx = PAdx = PdV
Where F is force and Adx is the infinitesimal change in volume.
You can now easily see that:
nC_vdT = -PdV
Second: First, remember that:
C_p = C_v + R
For convenience, let's adopt that:
C_p/C_v = \gamma
Using the famous equation, PV=nRT in the expression we derived in the first question:
nC_vdT = \frac{-nRT}{V}dV
\frac{dT}{T}+\frac{R}{C_v}\frac{dV}{V} = 0
And:
\frac{dT}{T}+(\gamma-1)\frac{dV}{V} = 0
Integrating:
\int \frac{dT}{T} dT + (\gamma-1)\int \frac{dV}{V} = 0
Which results in:
\ln (TV)^{(\gamma-1)} = cte
And finally:
(TV)^{(\gamma-1)} = cte
For a initial state (T_1, V_1) and a final state (T_2,V_2):
\frac{V_1}{V_2}^{(\gamma-1)} = \frac{T_2}{T_1}
Third: Try to solve the last one for yourself. Now it's easy ;)

 

[krootox217]

Thanks for the answers, I try to understand it and if I don't get everything, I ask again if that's ok :)