Question

The compressibility factor for one mole of a van der waals gas at 0c and 100 atm pressure is found to be 0.5. Assuming that the volume of the gas molecules is negligible, the van der waals constant a will be

Answer 1

Answer:

Compressibility factor, Z = 0.5

No. of moles, n = 1 mole

Pressure, P = 100 atm

Temperature, T = 0℃ = 273 K

R = 0.082 L atm K⁻¹ mol⁻¹

We know, that the compressibility factor is given as,

Z = PV / nRT …… (i)

⇒ 0.5 = [100 * V] / [1 * 0.082 * 273]

⇒ V = 0.1119 L

The Van der Waals equation is given as,

[P + a * (n/V)²] [(V/n - b)] = RT ….. (ii)

It is given that the volume of the gas molecules i.e., “b” is negligible, then eq. (ii) for 1 mole of gas becomes,

[P + a*(1/V)²] [(V/1 - 0)] = RT

⇒ [P + a/V^2][V] = RT

⇒ [(P/VRT) + (a/VRT)] = 1

⇒ a = [1 - Z] VRT

⇒ a = [1 – 0.5] * 0.1119 * 0.082 * 273

⇒ a = 1.252 atm L² mol⁻²

Hence, the Van der Waal's constant “a” will be 1.252 atm L² mol⁻² .

Answer 2

Explanation:

by using pv=nrt...........