Answer
a)$[10] = \frac{m^6.kPa}{kmol^2}$
b)$W = 403.3\ kJ$
Work Step by Step
a) The left-hand side of the equation expands to: $\bar{v}P + 10/\bar{v}$, therefore these two terms must have the same units to make the addition operation possible.
$(m³/kmol).kPa = \frac{[10]}{m³/kmol}$
$[10] = \frac{m^6.kPa}{kmol^2}$
b) Re-arranging the equation of state we get to:
$P = \frac{R_uT}{\bar{v}}-\frac{10}{\bar{v}^2}$
$P = \frac{nR_uT}{V}-\frac{10n^2}{V^2}$
With the work equation being:
$W = \int PdV$
$W = \int_{2\ m³}^{4\ m³} \frac{nR_uT}{V}-\frac{10n^2}{V^2} dV$
$W = (nR_uT\ln(V) +\frac{10n^2}{V})|_{2\ m³}^{4\ m³}$
Given the values of:
$n = 0.2\ kmol,\ R_u=8.314kJ/kmol.K,\ T=350K$
$W = 403.3\ kJ$
