For one mole of any van der Waal's gas $a = 0.27 atm-litre 2 / {mol}^{2}$ and $b = 1.218 {cm}^{3} / mol$ Select the correct statements (s).

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At $600^{\circ} C$ and $680 atm$ it will exist in gaseous form.

At $2700 K$ and low pressure, compression factor $^{'} Z^{'} > 1$

At $400^{\circ} C$ and $6800 atm$ , it will exist in liquid form.

At $2700 K$ and low pressure, compression factor $^{'} Z^{'} < 1$

answer is A, B.

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Detailed Solution

$\begin{array}{l} T_{C} = \frac{8 a}{27 R b} = \frac{8 \times 0.27}{27 \times 0.0821 \times 1.218 \times 10^{- 3}} \\ = 800 K = 527^{\circ} C \\ P_{C} = \frac{a}{27 b^{2}} = \frac{0.27}{27 {(1.218 \times 10^{- 3})}^{2}} = 6740.7 atm \\ T_{B} = \frac{a}{R b} = \frac{0.27}{0.0821 \times 1.218 \times 10^{- 3}} = 2700 K \end{array}$