A gas has critical pressure $33.5 atm$ and van der Waals constant $b = 0.0391 L mo l^{- 1}$ . Using $P_{c} = a /27 b^{2}$ and $T_{c} = 8 a /27 R b$ with $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , its critical temperature in K (one decimal) is

Numerical answer type, answer in K, tolerance 0.3.

Answer: 127.6 K

Dividing the two relations,

T_{c} / P_{c} = 8 b / R

, so

T_{c} = \frac{8 P _{c} b}{R} = \frac{8 \times 33.5 \times 0.0391}{0.0821} = \frac{10.4788}{0.0821} = 127.63 K

, i.e.

127.6 K