States of Matter 1 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 $11.0 \mathrm{~atm}$ at a temperature of $300 \mathrm{~K}$. The value of $b$ is $0.05 \mathrm{~L} \mathrm{~mol}^{-1}$.
(1998, 4M)
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Answer:
Correct Answer: 73. $(0.14)$
Solution:
- The van der Waals’ equation is
$$ \left(p+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T $$
$$ \begin{aligned} \Rightarrow \quad a=\frac{V^{2}}{n^{2}}\left[\frac{n R T}{V-n b}-p\right] & =\frac{(4)^{2}}{(2)^{2}}\left[\frac{2 \times 0.082 \times 300}{4-2(0.05)}-11\right] \ & =6.46 \mathrm{~atm} \mathrm{~L}^{2} \mathrm{~mol}^{-2} \end{aligned} $$
