A gas has $P_{c} = 51.0 atm$ and $V_{c} = 74.8 mL mo l^{- 1}$ . If it behaves as a van der Waals gas ( $Z_{c} = 3/8$ ), its critical temperature in K (one decimal, $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ ) is

Numerical answer type, answer in K, tolerance 0.3.

Answer: 123.9 K

\frac{P _{c} V _{c}}{R T _{c}} = \frac{3}{8}

, so

T_{c} = \frac{8 P _{c} V _{c}}{3 R} = \frac{8 \times 51.0 \times 0.0748}{3 \times 0.0821} = \frac{30.5184}{0.2463} = 123.91 K

, i.e.

123.9 K