Two moles of a real gas confined in a 5L flask exerts a pressure 9.1 a

1.	Two moles of a real gas confined in a 5L flask exerts a pressure 9.1 atm at a temperature of 27ºC. Calculate the value of 'a. The value of 'b' is 0.052 L mol-1 .
Answer» Given, n = 2 mol, V = 5L, P = 9.1 atm, T = 273 + 27 = 300K R = 0.082 L atm mol^-1 K^-1 According to van der Waals equation, \(\Big(P+\frac{an^2}{V^2}\Big)\)\((V-nb)\) = nRT Or a = \(\Big(\frac{nRT}{V-nb}-P\Big)\) x \(\frac{V^2}{n^2}\) = \(\Big(\frac{2\times0.082\times300}{5-2\times0.052}-9.1\Big)\)x \(\frac{(5)^2}{(2)^2}\) = \(\Big(\frac{49.2}{4.896}-9.1\Big)\)x \(\frac{25}{4}\) = (10.05 - 9.1) x \(\frac{25}{4}\) = 0.95 x \(\frac{25}{4}\) = = 5.938 atm L² mol^-2

Two moles of a real gas confined in a 5L flask exerts a pressure 9.1 atm at a temperature of 27ºC. Calculate the value of 'a. The value of 'b' is 0.052 L mol-1 .

Answer»

Given, n = 2 mol, V = 5L, P = 9.1 atm,

T = 273 + 27 = 300K

R = 0.082 L atm mol^-1 K^-1

According to van der Waals equation,

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

Or a = \(\Big(\frac{nRT}{V-nb}-P\Big)\) x \(\frac{V^2}{n^2}\)

= \(\Big(\frac{2\times0.082\times300}{5-2\times0.052}-9.1\Big)\)x \(\frac{(5)^2}{(2)^2}\)

= \(\Big(\frac{49.2}{4.896}-9.1\Big)\)x \(\frac{25}{4}\)

= (10.05 - 9.1) x \(\frac{25}{4}\)

= 0.95 x \(\frac{25}{4}\)

= = 5.938 atm L² mol^-2