Question

Using van der Waals equation, calculate the constant ${}^{\prime }{a}^{\prime }$ when 2 moles of a gas confined in a 4 liter flask exerts a pressure of 11.0 atm at a temperature of 300K. The value of ${}^{\prime }{b}^{\prime }$ is 0.05 litre ${mol}^{-1}$

Answer 1

van der Waals equation for n moles of gas is:

[P + $\frac{n^{2}a}{V^2}$ ] [V - nb] = nRT

Given V = 4 litre, P = 11.0 atm, T = 300K

b = 0.05 litre $mo{l}^{-}$${}^1$, n = 2

So,

[11 + $\frac{2^{2}a}{4^2}$ ] $[4-2\times 0.05]=nRT$

= $2\times 0.082\times 300$

a = 6.46 atm $litr{e}^{2}mo{l}^{-}$${}^2$