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Thermodyanmics, basic Integration and total differential.

  • Thread starter binbagsss
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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 :) !!
 

Answers and Replies

  • #2
HallsofIvy
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This problem is more physics than mathematics but:
- 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....?)
No. That equation is always true, not just for reversible reactions.

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?
As you add later, you are given that T is a constant.
[tex]\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[/tex]
 

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