Question

1 mole of monoatomic gas confined in a 22.5 litre flask at 273 K exerts a pressure of 0.98 atm whereas expected pressure is 1 atm. Has the gas behaved ideally? determine a, b and boyle's temperature.

Answer 1

Dear Student,

At temperature = 273 K, 1 mole of gas should exert a pressure of 1 atm and volume of 22.4 L. But it is exerting a pressure of 0.98 atm and has volume of 22.5 L. So, the gas is not behaving ideally.
For real gases of 1 mole,
$(P+\frac{a}{V^2})(V-b)=RT$
$$\begin{gathered} (P+\frac{a}{V^2})=1.0 \\ (0.98+\frac{a}{V^2})=1.0 \\ \frac{a}{V^2}=0.02 \\ \frac{a}{22.5^2}=0.02 \end{gathered}$$
a = 10.125
V-b = 22.4
22.5-b = 22.4
b = 0.1
Boyles temperature = T_b = $\frac{a}{Rb}=\frac{10.125}{0.082\times 0.1}$= 1234 K
Boyle's temperature = 1234 K

Regards