The ideal gas equation p V = n R T implies that the compressibility factor Z = p V / n R T = 1 Howev — Physical Chemistry Chemistry Question
Van der Waals gases
The ideal gas equation p V = n R T implies that the compressibility factor
Z = p V / n R T = 1
However, the compressibility factor is known to deviate from 1 for real gases. In order to account for the behavior of real gases, van der Waals proposed the following equation of state :
(p + n^2 a / V^2) (V - nb) = n R T
where a and b are constants, characteristic of the gas. The constant a is a measure of the intermolecular force and b that of the size of the molecules.
Show on the basis of van der Waals equation that
i. at sufficiently high temperatures Z is greater than unity for all pressures. At high temperatures and low pressures, Z approaches the value for an ideal gas.
ii. at lower temperatures, Z can be less than unity.
iii. for a = 0, Z increases linearly with pressure.
Model Answer
For a van der Waals gas
Z = pV / nRT = V / (V - nb) - na / VRT = 1 / (1 - nb/V) - na / VRT
The ratio of the magnitudes of the second and third terms on the right side is:
b / a R T (V / n), taking p V = n R T up to zeroth order.
The ratio of the magnitudes of the fourth and third terms on the right side is :
bn / V (V / RT) = bp / RT
i. From the ratios above, it follows that at sufficiently high temperature for any given pressure, the second term dominates the third and fourth terms. Therefore,
Z = 1 + bp / RT > 1
For small p, Z nearly equals to one.
ii. At lower temperatures, the third term can be greater (in magnitude) than the second term. It may be greater (in magnitude) than the fourth term also, provided p is not too large. Since the third term has a negative sign, this implies that Z can be less than unity.
iii. For a = 0
Z = 1 + bp / R T
which shows that Z increases linearly with p.
At a certain temperature, the variation of Z with P for He and N2 is shown schema-tically in the following figure.
For He, a = 3.46×10–2 bar dm6 mol-2 and b = 2.38×10–2 dm3 mol-1
For N2, a = 1.37×10–2 bar dm6 mol-2 and b = 3.87×10–2 dm3 mol-1
Identify the graph corresponding to He and N2.
Model Answer
Helium has negligible value of a. Graph (1) corresponds to He and (2) corresponds to N2.
Two p-V isotherms of a van der Waals gas are shown below schematically.
Identify the one that corresponds to a temperature lower than the critical temperature (Tc) of the gas.
Model Answer
Above T > Tc, only one phase (the gaseous phase) exists, that is the cubic equation in V has only one real root. Thus isotherm (2) corresponds to T < Tc .
For a given P, the three roots of van der Waals equation in V coincide at a certain temperature T = Tc. Determine Tc in terms of a and b, and use the result to show that N2 is liquefied more readily than He.
Model Answer
At T = Tc the three roots concide at V = Vc . This is an inflexion point.
dp / dV = 0 and d^2p / dV^2 = 0
The first condition gives
RTc / (Vc - nb)^2 = 2an^2 / Vc^3 (1)
The second condition gives
2RTc / (Vc - nb)^3 = 6an^2 / Vc^4 (2)
These equations give
Vc = 3 n b and Tc = 8a / 27bR
For He, Tc = 5.2 K
For N2, Tc = 128 K
Since Tc (N2) is greater than Tc (He), N2 is liquefied more readily than He.
Determine the work done by 1 mol of N2 gas when it expands reversibly and isothermally at 300 K from 1.00 dm3 to 10.0 dm3, treating it as a van der Waals gas.
Model Answer
W = ∫ p dV
= ∫ (RT / (V - b) - a / V^2) dV
= RT ln((V2 - b) / (V1 - b)) + a(1/V2 - 1/V1)
= 56.7 dm3 bar mol-1