Question
Question: For \(CO\), isotherm is of the type as shown. Near the point compressibility factor \(Z\) is? 
2.(1−Vb)
3.(1+RTVa)
4.(1−RTVa)
Solution
This question gives the knowledge about the compressibility factor and van der Waals gas equation. Compressibility factor is the factor which is used for transforming the ideal gas law to justify the behavior of real gases. It is also known as the gas deviation factor.
Formula used: The formula used to determine the compressibility factor for van der Waals gas equation is as follows:
(P+V2a)(V−b)=RT
Where P is the pressure, Vis the volume, Ris the gas constants, T is the temperature, a and b are van der Waals gas constants.
Complete step-by-step answer:
Van der Waals gas equation is the equation which corrects for the two properties of real gases: attractive forces between the gas molecules and the excluded volume
of gas particles.
Consider the van der Waals gas equation as follows:
⇒(P+V2a)(V−b)=RT
At very low pressure P , volume Vis very high.
V−b≈V
Substitute V−b as V in the van der Waals gas equation.
⇒(P+V2a)V=RT
On simplifying the above equation, we get
⇒PV+Va=RT
On further simplifying, we have
⇒PV=RT−Va
Divide the above equation with RT,
⇒RTPV=RTRT−VRTa
On further simplifying, we have
⇒RTPV=1−VRTa
Consider this as equation 1.
As we know,
Compressibility factor is the factor which is used for transforming the ideal gas law to justify the behavior of real gases. It is also known as the gas deviation factor. It is generally represented by Z.
Z=RTPV
Now, substitute RTPV as Z in equation 1 as follows:
⇒Z=1−VRTa
Therefore, the compressibility factor Z is 1−VRTa.
Hence, option 4 is correct.
Note: Van der Waals gas equation results in the correction of the two properties of real gases, one of which is attractive forces between the gas molecules and the second is the excluded volume of gaseous particles.