JEE Main · 2014 · 06 Apr 2014 · Offline · States of Matter

If Z is a compressibility factor, van der Waals equation at low pressure can be written as:

(A) $Z = 1 - \frac{Pb}{RT}$
(B) $Z = 1 + \frac{Pb}{RT}$
(C) $Z = 1 + \frac{RT}{Pb}$
(D) $Z = 1 - \frac{a}{VRT}$ ✓ Correct

Answer: (D)

Solution

Start from the van der Waals equation for one mole,

(P + \frac{a}{V ^{2}}) (V - b) = R T

, and expand:

P V = R T + P b - \frac{a}{V} + \frac{ab}{V ^{2}}

. At low pressure the

P b

and

\frac{ab}{V ^{2}}

terms are tiny beside

R T

, leaving

P V = R T - \frac{a}{V}

. Dividing by

R T

, the compressibility factor is

Z = \frac{P V}{R T} = 1 - \frac{a}{V R T}

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option

Practise more: all JEE Main PYQs on States of Matter →

More JEE Main questions on States of Matter

At the sea level, the dry air mass percentage composition is given as nitrogen gas : 70.0 , oxygen gas : 27.0 and argon gas : 3.0. If total pressure is 1.15 atm , then calculate the ratio of followings respectively :

(i) partial pressure of nitrogen gas to partial pressure of ox …
JEE Main 2025View solution →
Arrange the following gases in increasing order of van der Waals constant 'a'
A. Ar
B. $CH_{4}$
C. $H_{2} O$
D. $C_{6} H_{6}$

Choose the correct option from the following :-
JEE Main 2023View solution →
Three bulbs are filled with $CH X_{4}$ , $CO X_{2}$ and Ne as shown in the picture. The bulbs are connected through pipes of zero volume. When the stopcocks are opened and the temperature is kept constant throughout, the pressure of the system is found to be ______ atm. (Nearest i …
JEE Main 2023View solution →
At constant temperature a gas is at a pressure of 940.3 mm Hg. The pressure at which its volume decreases by 40% is ____ mm Hg.
(Nearest Integer)
JEE Main 2023View solution →
At 600 K, the root mean square (rms) speed of gas X (molar mass $= 40$ ) is equal to the most probable speed of gas $Y$ at 90 K. The molar mass of the gas Y is ______ $g mol^{- 1}$ . (Nearest integer)
JEE Main 2023View solution →
A certain quantity of real gas occupies a volume of $0.15 dm^{3}$ at 100 atm and 500 K when its compressibility factor is 1.07. Its volume at 300 atm and 300 K (When its compressibility factor is 1.4) is $____$ $\times 1 0^{- 4} dm^{3}$ (Nearest integer)
JEE Main 2023View solution →