Question

If the temperature of the gas is lower than Boyle's temperature, $T_B$, then

• A. At very low pressure $Z$ decreases with pressure
• B. At the very low pressure $Z$ increases with pressure
• C. At the very high pressure $Z$ decreases with pressure
• D. $Z$ becomes independent of the pressure.

Answer 1

The correct option is A At very low pressure $Z$ decreases with pressure

$(P+\frac{a}{V_m^2})(V_m-b)=RT$
$\Rightarrow \frac{P{V}_m}{RT}=Z=\frac{V_m}{(V_m-b)}-\frac{a}{RT{V}_m}$
$\Rightarrow Z={(1-\frac{b}{V_m})}^{-1}-\frac{a}{RT{V}_m}$
$\Rightarrow Z=1+\frac{b}{V_m}+\frac{b^2}{V_m^2}+...-\frac{a}{RT{V}_m}$
$\Rightarrow Z=1+(b-\frac{a}{RT})\frac{1}{V_m}+\frac{b^2}{V_m^2}+...$
Neglecting the higher terms.
$Z=1+(b-\frac{a}{RT})\frac{1}{V_m}$
$\frac{P{V}_m}{RT}=Zor\frac{1}{V_m}=\frac{P}{ZRT}$
$\Rightarrow Z=1+(b-\frac{a}{RT})\frac{P}{ZRT}$
$\Rightarrow Z(Z-1)=(b-\frac{a}{RT})\frac{P}{RT}$
$\Rightarrow Z^2-Z=(b-\frac{a}{RT})\frac{P}{RT}$
On differentiating with respect to $P$ we have
$2Z\frac{dZ}{dP}-\frac{dZ}{dP}=(b-\frac{a}{RT})\frac{1}{RT}$
$\Rightarrow \frac{dZ}{dP}=\frac{1}{(2Z-1)}\times (b-\frac{a}{RT})\frac{1}{RT}$
$\therefore {lim}_{P\to 0}\frac{dZ}{dp}=(b-\frac{a}{RT})\frac{1}{RT}\text{(As}P\to 0,Z=1$
$\text{When}T<T_B\text{then}T<\frac{a}{Rb}$ or $b<\frac{a}{RT}$
Hence, $b-\frac{a}{RT}$ is negative therefore ${lim}_{p\to 0}\frac{dz}{dp}$ is negative - it means at very low pressure Z decreases with pressure below the Boyle's temperature.