Question

The compression factor (compressibility factor) for 1 mole of
a van der Waal’s gas at 0°C and 100 atm pressure if found to
be 0.5. Assuming that the volume of gas molecules is
negligible, calculate the van der Waal’s constant 'a'.
(a) 0.253 L²mol⁻²atm (b) 0.53 L²mol⁻²atm
(c) 1.853 L²mol⁻²atm (d) 1.253 L²mol⁻²atm

Answer 1

Answer:

Option b is the correct g

Answer 2

The Van der Waal's constant 'a' is :

•The compression factor (compressibility factor) for 1 mole of a van der Waal’s gas at 0°C and 100 atm pressure if found to be 0.5.

• Assuming that the volume of gas molecules is negligible

• Given , n = 1, T = 0°C = 273K, P = 100 atm and Z = 0.5

• Z = PV / RT

0.5 = 100×V / 0.0821×273

V = 0.5×0.0821×273/100

V = 0.112 L

• Van der Waal's equation for real gas is given as

( P + na^2/V^2)(V - nb) = RT

For 1 mole, ( P + a^2/V^2)(V - b) = RT

• But we know that, volume of gas molecules is negligible. Therefore b = 0

( P + a^2/V^2)×V = RT ....( 1 )

• Substituting value of P, V, R and T in ( 1 )

[ 100 + a^2/(0.112)^2 ]×(0.112) = 0.0821×273

• Calculate it by yourself,

You'll get a = 1.253 L^2 mol^-2 atm