The work required to compress a gas is given by the following integral:
$$
\mathrm{W}=\int_{\mathrm{V}_1}^{\mathrm{V}_1} \mathrm{PdV}
$$
Van der Waals equation of state for a gas is is follows:
$$
\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}-\mathrm{nb}}-\frac{\mathrm{n}^2 \mathrm{a}}{\mathrm{V}^2}
$$
$\mathrm{R}=82.06\left(\mathrm{~cm}^3 \cong \mathrm{atm}\right) /(\mathrm{g}$ mole $\cong \mathrm{K}), \mathrm{b}=36.6 \mathrm{~cm}^3 / \mathrm{g}$ mole, and $\mathrm{a}=1.33 \times 10^6$ (atm $\cong$ $\left.\mathrm{cm}^6\right) /(\mathrm{g}$ mole $) 2$, when $\mathrm{P}$ is in atm, $\mathrm{T}$ in Kelvin, and $\mathrm{V}$ in $\mathrm{cm}^3$. Calculate the work done when the pressure of the gas is increased from 1 to $10 \mathrm{~atm}$. There was originally I liter of gas at $293 \mathrm{~K}$.