Question:

If pressure of $CO_2$ (real gas) in a container is given by $p = \frac{RT}{2V - b} - \frac{9}{4b^2},$ then mass of the gas in container ts

Updated On: Jul 6, 2022
  • 11 g
  • 22 g
  • 33 g
  • 44 g
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The Correct Option is B

Solution and Explanation

van der Waals' gas equation for $\mu$ mole of real gas $ \, \, \, \, \, \, \, \, \bigg( p + \frac{\mu^2 a}{V^2}\bigg) (V - \mu b) = \mu \, RT$ $ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, p = \bigg( \frac{\mu RT}{V - \mu b}\bigg) - \frac{\mu^2 a}{V^2}$ Given equation, $ \, \, \, \, \, \, \, \, \, \, p = \bigg( \frac{RT}{2V - b} - \frac{a}{4b^2}\bigg)$ On comparing the given equation with this standard equation, we get $ \, \, \, \, \, \, \, \mu = \frac{1}{2}$ Hence, $\mu = \frac{m}{M} \Rightarrow $ mass of gas $ \, \, \, \, \, \, \, \, \, \, \, m = \mu M = \frac{1}{2} \times 44 = 22g$
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