Question

the critical temperature of a gas if it gets liquified at 4.1 atm and has a critical volume of 3L

Answer 1

Answer:

73.89 atm

also PC=27b2a=73.89 atm

Critical temperatureTC=27Rb8a=300k=27oC

Taking ratio, we get

73.89300=R8b

⇒b=0.041

Now, b=4× volume occupied by 1 mole

∴ in volume of 1 mole=4b

∴ Volume of 24 moles=24×4b

=6b

=6×0.04161

=0.24966

=0.25L

or 250 mL.

Explanation:

mark me as brilliant