States of Matter - Result Question 73
73. Using van der Waals’ equation, calculate the constant $a$ when two moles of a gas confined in a four litre flask exert a pressure of $11.0 atm$ at a temperature of $300 K$. The value of $b$ is $0.05 \hspace {1mm} L \hspace {1mm} mol^{-1}$.
(1998, 4M)
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Answer:
Correct Answer: 73. (6.46)
Solution:
The van der Waals’ equation is
$ \left(p+\dfrac{n^{2} a}{V^{2}}\right)(V-n b)=n R T $
$\Rightarrow \quad a=\dfrac{V^{2}}{n^{2}}\left[\dfrac{n R T}{V-n b}-p\right] =\dfrac{(4)^{2}}{(2)^{2}}\left[\dfrac{2 \times 0.082 \times 300}{4-2(0.05)}-11\right] $
$ =6.46 \hspace {1mm} atm \hspace {1mm} L^{2} \hspace {1mm} mol^{-2}$
BETA
