Calculate the critical temperature of a Van der Waals gas for which $P_{c} = 100 atm$ and b is $0.050 {dm}^{3} / mol$ .
Take $R = \frac{1}{12} l atm/mol K$

300

K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

480

K

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

420

K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

400

K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 480 $K$
$P_{c} = 100 atm b = 0.050 {dm}^{3} / mol T_{c} = \frac{8 a}{27 R b} a = 3 P_{c} V_{c}^{2} b = \frac{V_{c}}{3} V_{c} = 3 b a = 27 P_{c} b^{2} T_{c} = \frac{8 \times 27 P_{c} b^{2}}{27 R b} T_{c} = \frac{8 P_{c} b}{R} T_{c} = \frac{8 \times 100 \times 1 \times 0.05}{1 / 12} = 480 K$

Suggest Corrections

Similar questions

The value of critical temperature $T_{C}$ for a gas is :

b = 0.04281 L m o l e^{- 1} .

V_{c} = 3 b, P_{c} = \frac{a}{27 b^{2}}

T_{c} = \frac{8_{a}}{27 b R}

0.40 g / {c m}^{3}

T_{C} = \frac{2 \times 10^{5}}{821} K

a

L^{2} {m o l}^{- 2}

Join BYJU'S Learning Program