Home
Class 12
CHEMISTRY
Under critical states of a gas for one m...

Under critical states of a gas for one mole of a gas, compressibility factor is :

A

`3//8`

B

`8//3`

C

`1`

D

`1//4`

Text Solution

Verified by Experts

The correct Answer is:
A

For `1mol` of gas `Z = (P_(C)V_(C))/(RT_(C))` (Under critical condition)
But, `P_(C) = (a)/(27b^(2)) , V_(C) = 3b, T_(C) = (8a)/(27Rb)`
`:. Z = ((a)/(27b^(2))) xx (2b)/(R ) xx (27Rb)/(8a) = (3)/(8)`
Hence, (A)`.
Promotional Banner

Topper's Solved these Questions

  • GASEOUS STATE

    RESONANCE|Exercise Exercise-1|1 Videos
  • GASEOUS STATE

    RESONANCE|Exercise Exercise|73 Videos
  • GASEOUS STATE

    RESONANCE|Exercise Exercise-3|1 Videos
  • FUNDAMENTAL CONCEPT

    RESONANCE|Exercise ORGANIC CHEMISTRY(Fundamental Concept )|40 Videos
  • HYDROCARBON

    RESONANCE|Exercise ORGANIC CHEMISTRY(Hydrocarbon)|50 Videos

Similar Questions

Explore conceptually related problems

For an ideal gas, the value of compressibility factor is zero.

At critical temperature pressure and volume. The compressibility factor (Z) is-

Compressibility factor Z=(PV)/(RT) . Considering ideal gas, real gas, and gases at critical state, answer the following questions: The compressibility factor of a real gas is

For H_(2) gas, the compressibility factor,Z = PV // n RT is -