Question

Using van der Waals equation, find the constant ${}^{\prime }{a}^{\prime }$ (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 $0.05Lmo{l}^{-1}$. $(R=0.082atmL/Kmol)$ (as nearest integer)

Answer 1

$\begin{array}{c}\text{By Vanderwaal's Gas Equation}\\ (P+\frac{a{n}^2}{v^2})(v-nb)=nRT\end{array}$

$\begin{array}{c}\text{Given}P=11,n=2,V=4,b=0.05,R=0.082\\ T=300\end{array}$

$\begin{array}{c}(11+\frac{a}{4})\times (4-0.1)=2\times 0.082\times 300\\ 11+\frac{a}{4}=12.61\Rightarrow a=6.46\end{array}$

Answer is 6.