Question

If the critical temperature of a gas is 160 K, then its Boyle's temperature will be:

• A. 400 K
• B. 270 K
• C. 540 K
• D. 600 K

Answer 1

The correct option is C 540 K

We know, $T_c=\frac{8a}{27Rb}$
$T_b=\frac{a}{Rb}$
$\therefore \frac{T_c}{T_b}=\frac{\frac{8a}{27Rb}}{\frac{a}{Rb}}=\frac{8}{27}$
$\Rightarrow \frac{160}{T_b}=\frac{8}{27}$
$\Rightarrow T_b=20\times 27=540K$