Question

Using van der Waals' equation, find the constant $a$ (in $atm$ $L^{2}mo{l}^{-2}$) when two moles of a gas confined in $4L$ flask exerts a pressure of $11.0$ atmospheres at a temperature of $300K$. The value of $b$ is $0.05Lmo{l}^{-1}$. $(R=0.082atmL/Kmol)$

• A. $2.62$
• B. $2.64$
• C. $6.24$
• D. $6.46$

Answer 1

The correct option is D $6.46$

By using $(P+\frac{a{n}^2}{V^2})(V-nb)=nRT$ and putting values:
$(11+\frac{a{2}^2}{4^2})(4-0.1)=2\times 0.082\times 300,a=6.46$.