Thermodyanmics, basic Integration and total differential.

Example 1.

A non ideal gas undergoes isothermal doubling of the pressure and you are asked to calculate the total work done; where the equation of state V=v(T,p), has been specified. Where V is the volume, T the temperature.

Attempt:

- The only way around this that I can see is using W=-∫pdV. ( But I can only justify its use if the reaction is reversible, and we are not told this....?)
- Anyway, to continue with this method, I first of all find the total differential dV=∂V/∂pdp+∂V/∂T dT*
- The 2nd term will =0 as it is isothermal.
- Now my main quesstion is, before intergrating - dp - on the first term - surely I need to express T in terms of p before integrating? (whilst in the partial derivative it is treated as a constant, before I integrate it cant be correct to treat is as a constant as there is known information on the relationship between p and V)
E.g say I end up with 2po∫po k/T p^3 dP, where k is a constant, do I need to express T in terms of P before integrated?

On second thoughts, this seems okay to do this time around, as T is constant, but explicitly, where/how does this step in , such that I could generalise this for cases when T is not constant, see example below:

2nd Example

- To calculate the work done in a carnot cycle, ideal gas, let the volumes be Va,Vb,Vc,Vd, where Va-Vb is a isothermal expansion and it operates between reservoirs at Th and Tl.
- Okay, now if im to consider the work done in the adiabatic expansion:

Attempt

1) Use of W=-∫pdv is THIS TIME justified as the carnot cycle consists of reversible processes by definition.
- I attain the same chain rule as above *, this time used to attain the total differential of V=V(p,T) as related by the ideal gas law: pV=nRT
-W=∫=nR dT + ∫nRTP^-1dp
- the first term looks okay to deal with , ΔT must =Th-Tl (my justification being adiabatic expansion so the gas expands without any head input, it does work on the environment, so its temperature must decrease) so that's my limits; and 'dT' looks fine to carry out as nR is a constant.
- But now I look at the second term, T is not constant. Surely it must be expressed in terms of p before carrying out the integration. Well I can use pV=nRT , so attain T=T(P,V), so now I want to eliminate the V to get everything solely in terms of p, well I can turn to pV^γ =constant. (as the process is adiabatic)*2

- BUT, the question did not specify any information regarding gamma - Cp/Cv or whether it is monoatomic etc.

*2 Bit of a long shot, but could it be in someway that because the processes are reversible- equilibrium at all stages- that only the final and starting values of T are required.

Many Thanks in advance to anyone who can shed some light on this :) !!

HallsofIvy
$$\frac{k}{T}\int_{p_0}^{2p_0} p^3 dp= \frac{k}{4T}\left[p^4\right]_{p_0}^{2p_0}= \frac{k}{4t}(16p_0^4- p_0^4)= \frac{15k}{4T}p_0^4$$