Methane has $P_{c} = 45.8 atm$ and $V_{c} = 0.0990 L mo l^{- 1}$ . Using $a = 3 P_{c} V_{c}^{2}$ , 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 0.02.

Answer: 1.35 L^2 atm/mol^2

a = 3 P_{c} V_{c}^{2} = 3 \times 45.8 \times (0.0990)^{2} = 137.4 \times 0.009801 = 1.3467

, i.e.

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

. This is lower than the tabulated

2.25

because the van der Waals relations are only approximate.