A van der Waals fluid is placed in a cylinder of volume V0 at temperature T0. Th

Question

A van der Waals fluid is placed in a cylinder of volume V0 at temperature T0. The cylinder is 1/4 filled with liquid and 3/4 filled with vapor. The pressure is P1. A piston is now inserted into the cylinder until the system has been isothermally compressed to the cylinder are just completely filled with liquid.

The temperature is now raised a small amount to T1, and the piston withdrawn just sufficiently to keep the tiny bubble of vapor in existence. Express the small change in pressure P that accompanies this change in terms of the latent heat L and T0, T1, and V0.

Eq. needed dP/dT = L/T V

I pose this question again because the last one whoever answered is totally wrong, please read the question. i dont need you to prove the Eq is right, i need express the small change in pressure P that accompanies this change in terms of the latent heat L and T0, T1, and V0.

Explanation / Answer

At equilibrium,

u1- u2 =0

Here u is chemical potential

So,

dP/dT = - (d(u1- u2 ) /dT) / (d(u1- u2) /dP )

We know du1 /dT = -s1  and du2 /dT = -s2

du1 /dP = v1  and du2 /dT = v2

dP/dT = - (-s1 + s2) / (v1-v2)

and also g1 = h1-Ts1   and g2= h2-Ts2

At equilibrium g1=g2=0

dP/dT = - (-s1 + s2) / (v1-v2)

= (s1 - s2) / (v1-v2)

= (h1 - h2) / T(v1-v2)

= L/T(delta V)

Hence proved

Here 1 and 2 notations denote two different states.

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