Question

If the temperature of the gas is lesser than boyle's temperature, T_B, then –

• A. At the very low pressure the Z decreases with pressure
• B. At the very low pressure the Z increases with Pressure
• C. At the very high pressure the Z decreases with pressure
• D. Z become independent on the pressure

Answer 1

Answer: (A)

At the very low pressure the Z decreases with pressure

Answer 2

$\left( P + \frac{a}{V_{m}^{2}} \right)$ $\left( V_{m}–b \right)$ = RT

or $\frac{PV_{m}}{RT}$ = Z = $\frac{V_{m}}{(V_{m}–b)}$ –$\frac{a}{RTV_{m}}$

or Z = $\left( 1–\frac{b}{V_{m}} \right)^{–1}$ – $\frac{a}{RTV_{m}}$ = 1 + $\frac{b}{V_{m}}$ + $\frac{b^{2}}{V_{m}^{2}}$ + ……. – $\frac{a}{RTV_{m}}$

or Z = 1 + $\left( b–\frac{a}{RT} \right)$ $\frac{1}{V_{m}}$ + $\frac{b^{2}}{V_{m}^{2}}$ +….....

Neglecting the higher terms,

Z = 1 + $\left( b–\frac{a}{RT} \right)$ $\frac{1}{V_{m}}$ ; $\frac{PV_{m}}{RT}$ = Z

or $\frac{1}{V_{m}}$ = $\frac{P}{ZRT}$

or Z = 1 + $\left( b–\frac{a}{RT} \right)$ $\frac{P}{ZRT}$

or Z(Z – 1) = $\frac{P}{RT}$

or Z² – Z = $\left( b–\frac{a}{RT} \right)$ $\frac{P}{RT}$

or 2Z $\frac{dz}{dp}$ – $\frac{dz}{dp}$ = $\left( b–\frac{a}{RT} \right)$ $\frac{1}{RT}$

or $\frac{dz}{dp}$ = $\frac{1}{(2Z–1)}$ × $\left( b–\frac{a}{RT} \right)$ $\frac{1}{RT}$

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

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.