Question

Calculate the pressure in bar of 1350 mol krypton in a 175.0 L container at 298 K, if a = 2.325 L2barmol−2 and b = 0.0396 Lmol−1.

Answer 1

Ans :-

From Real gas equation :

$\frac{P+an^{2}}{V^{2}} (V-nb) = nRT$ .........................(1)

Here,

P = Pressure of Krypton = ?

a = Molecular interaction constant = 2.325 L² bar mol^-2 (given)

b = Volume correction constant = 0.0396 L mol^-1

n = Number of moles of Krypton = 1350 mol (given)

V = Volume = 175.0 L (given)

T = Temperature = 298 K (given) and

R = Universal gas constant = 8.314 x 10^-2 L.bar .K^-1.mol^-1

Now, From equation (1), we have

$\frac{P+(2.325)(1350)^{2}}{(175.0)^{2}} (175.0-(1350)(0.0396) = (1350)(0.08314)(298)$

$\frac{P + 4237312.5}{30625} = \frac{33447.222}{121.54}$

$P + 4237312.5 = 8427852.3$

$P = 8427852.3 - 4237312.5$

$P = 4190539.8 bar$

P = 4.19 x 10⁶ bar