Question

the compressibility factor for o2 is 0.308 and its critical pressure and critical volume are 50.1atm and 0.078litre/mol respectively. what wil be the boyle temperature of o2

Answer 1

It's answer is 428.38k   Hope that will help you.

Answer 2

Answer: The Boyle's temperature of $O_2$ is 428.64K.

Explanation: We are given:

$P_c=50.1atm\\\\V_c=0.078L/mol$

We know that,

$P_c=3b\\\\V_c=\frac{a}{27b^2}$

Now, calculating the value of 'a' and 'b'

$50.1atm=3b\\b=16.7atm$

$0.078L/mol=\frac{a}{27\times (16.7atm)^2}\\a=587.342Latm^2/mol$

The value of critical temperature is:

$T_c=\frac{a}{Rb}\\R=0.08205\text{ L atm }mol^{-1}K^{-1}$

Putting values of 'a' and 'b' in above equation, we get:

$T_c=\frac{587.342Latm^2/mol}{0.08205\text{ L atm }mol^{-1}K^{-1}\times 16.7atm}\\\\T_c=428.64K$