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:
[itex]dU = dQ-dW [/itex]
Because it's an adiabatic process, there is no heat exchange, so Q = 0.
[itex]dU = -dW [/itex]
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:
[itex]dU = nC_vdT [/itex]
As for work:
[itex]dW = Fdx = PAdx = PdV[/itex]
Where F is force and Adx is the infinitesimal change in volume.
You can now easily see that:
[itex]nC_vdT = -PdV[/itex]
Second: First, remember that:
[itex]C_p = C_v + R[/itex]
For convenience, let's adopt that:
[itex]C_p/C_v = \gamma[/itex]
Using the famous equation, PV=nRT in the expression we derived in the first question:
[itex]nC_vdT = \frac{-nRT}{V}dV[/itex]
[itex]\frac{dT}{T}+\frac{R}{C_v}\frac{dV}{V} = 0[/itex]
And:
[itex]\frac{dT}{T}+(\gamma-1)\frac{dV}{V} = 0 [/itex]
Integrating:
[itex]\int \frac{dT}{T} dT + (\gamma-1)\int \frac{dV}{V} = 0[/itex]
Which results in:
[itex]\ln (TV)^{(\gamma-1)} = cte[/itex]
And finally:
[itex](TV)^{(\gamma-1)} = cte[/itex]
For a initial state [itex](T_1, V_1)[/itex] and a final state [itex](T_2,V_2)[/itex]:
[itex]\frac{V_1}{V_2}^{(\gamma-1)} = \frac{T_2}{T_1}[/itex]
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 :)