Question

Which of the following is/are correct ?

• A. For the incompressible liquid $\left( \frac{dH}{dP} \right)_{T}$ is approximately equal to volume of liquid
• B. For ideal gas $\left( \frac{dH}{dP} \right)_{T}$ is equal to zero
• C. For real gas if $\left( \frac{dE}{dV} \right)_{T}$ = 0 then not necessarily $\left( \frac{dH}{dP} \right)_{T}$ is equal to zero
• D. All of the above are correct

Answer 1

Answer: (D)

All of the above are correct

Answer 2

H = E + PV

$\left( \frac{dH}{dP} \right)_{T}$ = $\left( \frac{dE}{dP} \right)_{T}$ + P $\left( \frac{dV}{dP} \right)_{T}$ + V

For liquid, $\left( \frac{dH}{dP} \right)_{T}$ = $\left( \frac{dE}{dV} \right)_{T}$ $\left( \frac{dV}{dP} \right)_{T}$ +

P $\left( \frac{dV}{dP} \right)_{T}$ + V

For incompressible liquid, $\left( \frac{dV}{dP} \right)_{T}$ 0.

\ $\left( \frac{dH}{dP} \right)_{T}$ V

For ideal gas, $\left( \frac{dH}{dP} \right)_{T}$ = 0

For the real gas, if $\left( \frac{dE}{dV} \right)_{T}$ = 0,

then $\left( \frac{dH}{dP} \right)_{T}$ = P $\left( \frac{dV}{dP} \right)_{T}$ + V ¹ 0

\ $\lim_{p \rightarrow o}$ $\frac{dz}{dp}$ $\overset{쳿}{–}$ $\left( b–\frac{a}{RT} \right)$ $\frac{1}{RT}$

When T < T_B then T < $\frac{a}{Rb}$ or b < $\frac{a}{RT}$

hence b – $\frac{a}{RT}$ is negative therefore $\frac{dz}{dp}$ is negative – it means at very low pressure Z decreases with pressure below the Boyle's temperature.