Question

The sketch shows the plot of Z vs P for 1 mole of a hypothetical gas at three distinct
temperatures:

Boyle's temperature is the temperature at which a gas shows ideal behaviour over a
pressure range in the low pressure region. Boyle's temperature $\left(T_b\right)=\frac{a}{Rb}$. If a plot is
obtained at temperatures well below Boyle's temperature, then the curve will show a negative deviation in the low pressure region and a positive deviation in the high pressure
region.

In the first figure, in the plot at $500K,Z$ changes from 2 to 2.2 as the pressure varies from 1000 atm to $1200\text{atm}$ (high pressure). The value of $\frac{b}{RT}$ will be:

• A. ${10}^{-3}at{m}^{-1}$
• B. $2\times {10}^{-3}at{m}^{-1}$
• C. $5\times {10}^{-4}at{m}^{-1}$
• D. ${10}^{-4}at{m}^{-1}$

Answer 1

The correct option is A ${10}^{-3}at{m}^{-1}$

At high pressure, $Z=1+\frac{Pb}{RT}$
$\therefore dZ=\frac{dP\times b}{RT}$
$\therefore 0.2=200\times \frac{b}{RT}$
$\Rightarrow \frac{b}{RT}=\frac{0.2}{200}={10}^{-3}at{m}^{-1}$