Question
Question: At low pressure, the Vander Waals equation is reduced to : (A) \(Z=\dfrac{p{{V}_{m}}}{RT}=1-\dfrac...
At low pressure, the Vander Waals equation is reduced to :
(A) Z=RTpVm=1−RTap
(B) Z=RTpVm=1−RTVa
(C) p(Vm−b)=RT
(D) Z=RTpVm=1+RTbp
Solution
Ideal gases obey the equation PV = nRT. However real gases do not obey the above law as they have pressure and volume correction terms added to the ideal pressure and volume. As we go on decreasing pressure, the value of pressure in the numerator becomes insignificant. On the other hand, the value of pressure tends to infinity when present in the denominator.
Complete step by step answer:
For ideal gases, Z = 1 as PV = nRT.
However for real gases we follow the Van der Waals equation i.e.
[P+V2an2][V−nb]=nRT
Where,
P stands for pressure,
V stands for volume,
n stands for number of moles,
a is pressure correction constant
b is volume correction constant
T stands for temperature
R stands for universal gas constant
At low pressure, the volume correction constant becomes negligible and the equation becomes,
(P + V2a) V = R . T
⇒ RTPV = ( 1 − RTVa ) = Z
Where,
Z is the compressibility factor.
Therefore, from the above calculation we can conclude that Z=RTpVm=1−RTVa and the correct answer is option (B).
Additional Information:
-Critical temperature is defined as the temperature of a substance in its critical state beyond which it cannot be liquefied.
-Critical pressure of a fluid is defined as the vapor pressure of the fluid at the critical temperature.
-Critical volume is the volume occupied by the fluid or substance in its critical state (Critical pressure and temperature). So, the correct answer is “Option B”.
Note: Compressibility factor (Z) is a correction factor which describes the deviation of a real gas from its behavior of an ideal gas. It is simply the ratio of molar volume of the gas to the molar volume of an ideal gas subjected to the same identical temperature and pressure.