One mole of a gas with $b = 0.0427 L mo l^{- 1}$ occupies $1.00 L$ at $273 K$ and exerts a pressure of $20.4 atm$ . Assuming the gas obeys the van der Waals equation ( $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ ), the value of $a$ in $L^{2} atm mo l^{- 2}$ (two decimals) is

Numerical answer type, answer in L^2 atm/mol^2, tolerance 0.03.

Answer: 3.01 L^2 atm/mol^2

Rearranging,

a = (\frac{RT}{V _{m} - b} - P) V_{m}^{2}

\frac{0.0821 \times 273}{1.00 - 0.0427} = \frac{22.413}{0.9573} = 23.413 atm

. Then

a = (23.413 - 20.4) \times 1.0 0^{2} = 3.013

, i.e.

3.01 L^{2} atm mo l^{- 2}