A substance whose state is specified by P, V, T can exist in...

A substance whose state is specified by $P, V, T$ can exist in two distinct phases. At a given temperature $T$ the two phases can coexist if the pressure is $P(T)$. The following information is known about the two phases. At the temperatures and pressures where they can coexist in equilibrium, (i) there is no difference in the specific volume of the two phases; (ii) there is no difference in the specific entropy of the two phases; (iii) the specific heat $c_p$ and the volume expansion coefficient $\alpha$ are different for the two phases. (a) Find $d P(T) / d T$ as a function of $T$. (b) What is the qualitative shape of the transition region in the $P-V$ diagram? In what way is it different from that of an ordinary gas-liquid transition? The phase transition we have described is a second-order transition.

A substance whose state is specified by $P, V, T$ can exist in two distinct phases. At a given temperature $T$ the two phases can coexist if the pressure is $P(T)$. The following information is known about the two phases. At the temperatures and pressures where they can coexist in equilibrium,
(i) there is no difference in the specific volume of the two phases;
(ii) there is no difference in the specific entropy of the two phases;
(iii) the specific heat $c_p$ and the volume expansion coefficient $\alpha$ are different for the two phases.
(a) Find $d P(T) / d T$ as a function of $T$.
(b) What is the qualitative shape of the transition region in the $P-V$ diagram? In what way is it different from that of an ordinary gas-liquid transition?
The phase transition we have described is a second-order transition.

Key Concepts

Phase Diagrams in Thermodynamics

Phase diagrams, such as the pressure–volume (P–V) diagram, visualize the states of a system and the transitions between them. For a second-order transition, the coexistence is marked not by a flat, two-phase region (as in first-order gas-liquid transitions) but by a smooth change in pressure with no discontinuous jump in volume. This leads to a transition region that appears as a continuous, often sharply curved, boundary rather than the constant pressure plateau typical of first-order transitions.

Ehrenfest Relations

Ehrenfest relations provide a method to relate the discontinuities in the second derivatives of the free energy to the slope of the coexistence curve in the pressure–temperature plane. They extend the idea of the Clausius–Clapeyron equation to second-order transitions, enabling one to determine the pressure derivative with respect to temperature even when the first order properties (such as volume and entropy) are continuous, by linking the differences in properties like specific heat and thermal expansion coefficient between the phases.

Second-Order Phase Transition

A second-order phase transition, as classified by Ehrenfest, is characterized by the continuity of the first derivatives of the thermodynamic potential (such as specific volume and entropy) with respect to temperature and pressure, while second derivatives (like specific heat and the volume expansion coefficient) are discontinuous. This means that there is no latent heat or volume change at the transition, but other response functions suddenly differ between the two phases.

Transcript

00:01 Hi everyone this is the problem based on the behavior of isotherm for real gas above and below the critical point.

00:43 Here it is given a positive slope del p .p upon dail v of pv graph signifies that an increase in pressure causes an increase in or decrease in volume results in decrease in pressure.

02:15 It cannot be the case of real guess.

02:41 Now see here is the second part.

02:49 So from above result we can conclude we cannot have a positive slope along an.

03:27 And isotherm and so can have no extremes along an isotherm when d 'l p upon d 'l v v v vans along an isotherm the point on the curve in pb diagram be an inflation point and second derivative of pressure with respect to volume must be zero c point according to gas equation for non -ideal gas partial derivative of pressure with respect to volume having the value nrt v minus nb whole square plus 2a n square by v q and second derivative would be 2n r t upon b minus n v q plus sorry it would be minus 6 a and a square b to the power 4 1 2 for equation 2 and 3 for equation 2 to v 0 and equation 2 2 v 0 we will get v q n r t plus 2 a n is square v minus n b whole square n b whole square n v to the power 4 n r t is equal to 3a n square v minus n b whole cube d if v is equal to 3 y 2 b y n b which is solved for b y n having the critical value 2 v 3b putting this value in equation 4 and 5 critical temperature we will get 8 a upon 27 rv e part critical pressure p c is equal to r t upon b upon e critical solving it, you will get rt critical is 8a upon 27 rv upon 2v minus a upon 9b squared.

10:23 So you will get a upon 27b square.

10:29 This is the critical pressure.

10:39 F r upon critical temperature upon vc into v .c upon v .c...

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