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