For methyl chloride vapor at $100C,$ second and third virial coefficients are:

$B=-242\mathrm{.}5c\frac{m^3}{m}ol$ and $C=25,200c\frac{m^6}{m}o{l}^2$

Calculate work of mechanically reversible, isothermal compression of $1$mol of methyl

chloride from $1$ bar to $55$ bar at $100C$ using:

a$.Z=1+\frac{B}{V}+\frac{C}{V^2}=P\frac{V}{R}T(10$ marks$)$

b$.Z=1+B^{\text{'}}P+C^{\text{'}}{P}^2(10$ marks$)$

$B^{\text{'}}=\frac{B}{RT}$ and $C^{\text{'}}=\frac{C-B^2}{(RT{)}^2}$

Also find $Z_1$ and $Z_2(5$ marks$)$

Hint for part $($a$)$:

I. First, find the volumes at both conditions $(1$ bar and $55$ bar$)$ by solving the cubic

equation for $V.$

II$.$ Then, use $W=-{\displaystyle \underset{V_1}{\overset{V_2}{}}}P.dV$ and substitute for $P$ in terms of $V$ from the given virial

equation of state

Hint for part $($b$)$:

I. You need to integrate $W$ in terms of $dP$ and not $dV($as was in part $($a$))$ since $Z$ here

is given in terms of $P.$

II$.$ Use $V=\frac{RTZ}{P}=\frac{RT}{P}(1+B^{\text{'}}P+C^{\text{'}}{P}^2),$ now differentiate this and find $dV$ in terms of

$P$ and $dP$

III. Substitute this dV into the work integral and find the work done from $P_1$ to $P_2.$

You cannot use direct IG relation for isothermal work because methyl chloride is not

ideal gas.