Test: Van der Waals Equation (Old NCERT) - JEE MCQ

For correction in intermolecular forces, we have ∆P = an²/V²
So the actual pressure becomes, P+an²/V²

More the value of a, more be easy to liquify the gas.
So, the order of ease of liquefaction is, NH₃ > CH₄ > N₂ > O₂

At high temperature and low pressure, the gas volume is infinitely large and both intermolecular force as well as molecular volume can be ignored. Under this condition postulates of kinetic theory applies appropriately and gas approaches ideal behavior.

Z = VObserved/VIdeal
If Z < 1
VObserevd>VIdeal
Or Vobseved>22.4 L
So, according to me, option a is  correct.

Compressibility factor of an ideal gas :

z = pV/nRT = 1

If a gas is deviated from its ideal behaviour, therefore, z >1 or 1 > z.

Hence, for positive deviation of a gas from ideal behaviour  z > 1.

At low pressure, volume correction for 1 mole of gas is negligible,
 b=0 and van der Waals equation becomes
[P+a/V²]V=RT
So, the correct option is A

`b` is equal to 4 times the volume of molecules in one mole of a gas N₀ molecules
Volumes of one molecule = v
Volume of N₀ molecule =vN₀
Hence, b=4vN₀

The compressibility factor for real gases at low pressure, high pressure and that of gases of low molar masses are z₁, z₂ and z₃.The values of z₁, z₂, z₃ can be derived as follows.
At low pressures, the gas equation can be written as,
(P + a/v2m) (Vm) = RT
or Z₁ = Vm / RT = 1 – a/VmRT
At higher pressure,
PV - Pb = RT
So, equating with real gas equation,
PV / RT = 1 + Pb/RT
Again, Z₂ = PV / RT
So, Z₂ = 1 + Pb / RT

Van der waals constant ′a′ is a mixture of inter molecular forces and ′b′ is a measure of molecular size. A greater value of ′a′ and lesser value of ′b′ supports qualification of a given gas. Therefore, chlorine is more easily liquefied than ethane because  
′a′ for Cl₂ > ′a′ for C₂H₆ but ′b′ for Cl₂ > ′b′ for C₂H₆.

The correct answer is option B
Vanderwaal's equation for 1 mole of gas is 
(P + a/V2 ​)( V − b ) = RT
But, b = 0......(given)
(P + a/V2​) (V) = RT
∴PV= (−a × 1/V​) + RT
We know the basic general equation of straight line, y= mx + c
Slope = tan( π − θ ) = −tan θ = −a
So, tanθ = a = 

‘R’ = J mol^-1K^-1
‘a’ = atm L² mol^-2 = bar (dm³)² mol^-2
‘b’ =  L mol^-1 = dm³ mol^-1

According to a real gas equation,
The constants 'a'  and 'b': Van der Waals constant for attraction 'a' and volume 'b' are characteristic constants for a given gas.
(i)The 'a' values for a given gas are a measure of intermolecular forces of attraction. More are the intermolecular forces of attraction, more will be the value of a.
(ii)For a given gas van der Waals constant of attraction 'a' is always greater than van der Waals constant of volume 'b'.
(iii)The gas having a higher value of 'a'  can be liquefied easily and therefore H2 and He is not liquefied easily.
According to this, for gas A (Z>1), a=0 and its dependence on P is linear at all pressure and for gas B (Z<1), b=0 and its dependence on P are linear at all pressure.
Also, at high pressure, the slope is positive for all real gases.

From the figure, their centres are separated by 2r at the point of contact
Volume of 1 molecule = 4/3 πr³ = V_mol
Effective volume = 4/3 π(2r)³ = 8 4/3πr³ = 8V_mol
Excluded volume per molecule = ½ 4/3 π(2r)³ = ½ 8 4/3πr³ = 4V_mol
Greater the excluded volume, greater the deviation from ideal behaviour.

The coefficient B in the virial equation on state PV_m=RT(1+B/V_m+ C/V_m²+...)
b.is equal to zero at Boyle’s temperature. At this temperature, the real gas behaves like ideal gas and the virial equation becomes ideal gas equation  PV_m=RT
c. has the dimension of molar volume. This is because in the virial equation, it appears in the form of the ratio B/V_m
d. B = b - a/RT is the relation b/w B aand van er walls constant.

Assertion is true. For real gas, pV vs p plot is not a straight line.
The reason for this is that for real gas, we have to consider the intermolecular forces which are ignored in ideal gas.
Reason is also correct.Z>1at higher  pressure for all gas but at low pressure, some gases have Z<1. However this is not the correct explanation of assertion. 

Vander waals constant 'a' is a measure of attractive force
The van der waals constant 'a' represents the magnitude of the attractive forces present between gas molecules. Higher is the value of 'a', more easily the gas can be liquefied.
However it has no relation with statement II. Statement II is a general statement.

Compressibility factor(Z) = PV/RT
For 1 mole of real gas, Van der Waals eqn is (P + a/V²)(V-b) = RT
At low pressure, volume is very large and hence correction term; b can be neglected in comparison to V
(P + a/V²)V= RT
PV + a/V = RT
PV/RT = 1- a/VRT
Z = 1-a/VRT

Above Z=1, we have Z>1 and below Z=1, we have Z<1. So, the correct answer be (1+b/V)

For H₂, we have always Z > 1. In the given graph, at position B we have Z > 1.

As we know, comprssibility factor (Z)