0

The first law applied to a liquid film is $dU = dQ +2\sigma dA$ where $\sigma$ stands for surface tension.

Or $dQ = dU + (-2\sigma) dA \leftrightarrow dQ = dU + PdV$

Now suppose we wish to find what happens to temperature if I increase the area adiabatically. (Throughout I assume the stretching to be reversible.)

i.e. we need to know $\left(\frac{dT}{dA}\right)_S$

I propose two methods here to calculate this.

Method 1 (quite straight-forward)

Clearly for adiabatic processes $dQ=0$ i.e. $dU = 2\sigma dA$. Here $dU=C_A dT$. So, we get $dT = \frac{2\sigma}{C_A} dA$

Or $\left(\frac{dT}{dV}\right)_S = \frac{2\sigma}{C_A}$

Method 2 (uses Maxwell relations)

As we know following Maxwell relation, $\left(\frac{dT}{dV}\right)_S = - \left(\frac{dP}{dS}\right)_V$

which in our case would correspond to $\left(\frac{dT}{dA}\right)_S = 2 \left(\frac{d\sigma}{dS}\right)_A = 2T \left(\frac{d\sigma}{dQ}\right)_A$.

Also $(dQ)_A = (dU)_A = C_A dT$

This would mean $\left(\frac{dT}{dA}\right)_S = \frac{2T}{C_A} \left(\frac{d\sigma}{dT}\right)_A$

My teacher says that it is the second method which is correct, but I can not figure out why first one is wrong since both can not certainly be right together.

The Imp
  • 1,036

2 Answers2

2

Your error in method 1 was assuming the U = U(T) and not U = U(T,A). But, you would have been much better off starting out directly with entropy S = S(T,A), so that $$dS=\frac{C_A}{T}dT+\left(\frac{\partial S}{\partial A}\right)_TdA$$Then from the Maxwell relationship, $$\left(\frac{\partial S}{\partial A}\right)_T=-2\left(\frac{\partial \sigma}{\partial T}\right)_A$$ So,$$dS=\frac{C_A}{T}dT-2\left(\frac{\partial \sigma}{\partial T}\right)_AdA$$

Chet Miller
  • 35,124
2

In the method 1, you expressed increase of energy during adiabatic process as

$$ dU=C_A dT $$

where $C_A$ is presumably capacity when the area $A$ is constant: $$ C_A = \left(\frac{dU}{dT}\right)_A. $$ But $C_A$ is not the right factor to use for adiabatic process, because in this process $A$ is not constant, but $S$ is. The energy increase formula is actually

$$ dU=C_S dT $$ where $$ C_S = \left(\frac{dU}{dT}\right)_S. $$

We then have the relation

$$ 2\sigma dA = \left(\frac{dU}{dT}\right)_S dT $$ But $C_S$ is not measurable, so we still cannot express sought quantity in terms of known things.

If we try to use Maxwell's relations on $C_S$, we find we can't, because the pair of variables $T,S$ is not sufficient to determine the thermodynamic state. Only pairs $(S,A), (T,A), (S, \sigma), (T,\sigma)$ qualify.

So the method 1 does not really work. The task is to express $dT/dA|S$ in terms of something known, and with the hint of the pair of variables $(S,A)$ in the low, one can try to use the Maxwell relation

$$ \frac{dT(S,A)}{dA} = 2\frac{d\sigma(S,A)}{dS}. $$

that follows from the fundamental relation

$$ dU = TdS +2\sigma dA. $$

Here $U,T,\sigma$ are all functions of $(S,A)$.

One then proceeds as in your method 2.

Ján Lalinský
  • 44,745