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Equation Thermodynamics


Scienc

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10 hours ago, Scienc said:

Why does the equation τ = -n.R.T.ln⁡ (v_2 / v_1) only work for reversible processes? 

Is that for isothermal expansion of an ideal gas?

Some of the setup looks different, negative sign and use of tau symbol. Could of course simply be rearranged though.

Edited by Endy0816
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13 hours ago, Scienc said:

Why does the equation τ = -n.R.T.ln⁡ (v_2 / v_1) only work for reversible processes? 

You are, as Endy0816 says, considering the work done by an ideal gas in an isothermal expansion or compression.

Following your notation,

\[\tau=-\int PdV=-\int P\left(V\right)dV\]

So that,

\[\tau=-\int nRT\frac{dV}{V}=-nRT\log\frac{V_{2}}{V_{1}}\]

The reason why that procedure is only valid for reversible processes is that if you want to be able to guarantee that the equation of state \[f\left(P,V,T,n\right)=\frac{PV}{nRT}=\textrm{constant}=1\] is valid throughout the process, it must take place under equilibrium conditions throughout. In other words, the control parameters must vary very slowly compared to the relaxation times of the gas, so that it constantly re-adapts to the "differentially shifted" equilibrium conditions. Those are called "reversible conditions". Chemists use the word "reversible" in a slightly different sense, so be careful if your context is chemistry.

6 hours ago, studiot said:

How would you define T in a non revesible process ?

Exactly.

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2 hours ago, joigus said:

Chemists use the word "reversible" in a slightly different sense, so be careful if your context is chemistry.

 

Like too the distinction between chemical thermo and physics thermo. That causes so much difficulty with students. +1

Edited by studiot
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