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In van der Waals equation at constant temperature $300 \mathrm{~K}$, if $a=1.4 \mathrm{~atm} \mathrm{~L} \mathrm{~mol}^{-2}$, $V=100 \mathrm{~mL}, n=1 \mathrm{~mole}$, then what is the pressure of the gas?
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The correct answer is:
106 atm
At moderate pressure van der Waals equation is given as :
$\begin{aligned} & \left(P+\frac{a n^2}{V^2}\right)(V)=n R T \\ & \left(P+\frac{1.4}{(0.1)^2}\right)(0.1)=1 \times 0.082 \times 300\end{aligned}$
$\begin{aligned} & (P+140) \times 0.1=24.6 \\ & 0.1 P+14=24.6 \\ & 0.1 P=10.6 \Rightarrow P=106 \mathrm{~atm}\end{aligned}$
$\begin{aligned} & \left(P+\frac{a n^2}{V^2}\right)(V)=n R T \\ & \left(P+\frac{1.4}{(0.1)^2}\right)(0.1)=1 \times 0.082 \times 300\end{aligned}$
$\begin{aligned} & (P+140) \times 0.1=24.6 \\ & 0.1 P+14=24.6 \\ & 0.1 P=10.6 \Rightarrow P=106 \mathrm{~atm}\end{aligned}$
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