⇒Thermodynamics

Correct Option is (A)

${\mathrm{W}}_{max}=-2.303\mathrm{n}\mathrm{R}\mathrm{T}{\log }_{10}\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

Hence, ${\mathrm{W}}_{max}\mathrm{\infty }\mathrm{n}$ (Given: $\mathrm{R},\mathrm{T},{\mathrm{V}}_2,{\mathrm{V}}_1=$ Constant)

$\therefore {\mathrm{W}}_{max}\mathrm{\infty }\frac{1}{\mathrm{M}.\mathrm{W}.}$ (Given: equal mass)

Hence, lower the molecular mass, greater is the work done. Among the given, ${\mathrm{H}}_2$ has the lowest molecular mass.

Correct Option is (D)

1 mole of ${\mathrm{C}}_2{\mathrm{H}}_4$ reacts with 1 mole of $\mathrm{HCl}$ to produce 1 mole of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$.

Hence, $150\mathrm{mL}$ of $\mathrm{HCl}$ would react with only $150\mathrm{mL}$ of ${\mathrm{C}}_2{\mathrm{H}}_4$ to produce $150\mathrm{mL}$ of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$.

${\mathrm{V}}_1=150\mathrm{mL}+150\mathrm{mL}=300\mathrm{mL}=0.3{\mathrm{dm}}^3$

${\mathrm{V}}_2=150\mathrm{mL}=0.15{\mathrm{dm}}^3$

$\mathrm{W}=-{\mathrm{P}}_{\text{ext }}\mathrm{\Delta }\mathrm{V}=-{\mathrm{P}}_{\text{ext }}({\mathrm{V}}_2-{\mathrm{V}}_1)$

$\therefore W=-1$ bar $(0.15{\mathrm{dm}}^3-0.3{\mathrm{dm}}^3)$

$\therefore \mathrm{W}=0.15{\mathrm{dm}}^3$ bar

$\therefore W=0.15{\mathrm{dm}}^3$ bar $\times 100\frac{\mathrm{J}}{{\mathrm{dm}}^3\text{ bar }}$

$\therefore \mathrm{W}=15\mathrm{J}$

The PV work in joules is $15\mathrm{J}$.

Correct Option is (C)

$\begin{array}{rl}\mathrm{\Delta }\mathrm{V}& =500{\mathrm{cm}}^3=0.5{\mathrm{dm}}^3\\ {\mathrm{P}}_{\mathrm{ext}}& =2\times {10}^5\mathrm{N}{\mathrm{m}}^{-2}=2\mathrm{bar}(\text{ since, }1\times {10}^5\mathrm{N}{\mathrm{m}}^{-2}=1\mathrm{bar})\\ \mathrm{W}& =-{\mathrm{P}}_{\mathrm{ext}}\times \mathrm{\Delta }\mathrm{V}\\ & =-2\times (0.5)\\ & =-1{\mathrm{dm}}^3\text{ bar }\\ \therefore \mathrm{W}& =-100\mathrm{J}(\text{ since, }1{\mathrm{dm}}^3\mathrm{bar}=100\mathrm{J})\end{array}$

According to first law of thermodynamics,

$\mathrm{\Delta }\mathrm{U}=\mathrm{Q}+\mathrm{W}=200\mathrm{J}-100\mathrm{J}=+100\mathrm{J}$

Correct Option is (D)

$\begin{array}{rl}& \mathrm{W}=-{\mathrm{P}}_{\text{ext }}\mathrm{\Delta }\mathrm{V}=-{\mathrm{P}}_{\text{ext }}({\mathrm{V}}_2-{\mathrm{V}}_1)\\ & {\mathrm{V}}_1=3{\mathrm{dm}}^3\\ & {\mathrm{V}}_2=?\\ & {\mathrm{P}}_{\text{ext }}=1.6\mathrm{bar}\\ & \mathrm{W}=-800\mathrm{J}=8{\mathrm{dm}}^3\text{ bar }(\because 100\mathrm{J}=1{\mathrm{dm}}^3\text{ bar })\\ & -8=-1.6({\mathrm{V}}_2-3)\\ & {\mathrm{V}}_2-3=5\\ & {\mathrm{V}}_2=8{\mathrm{dm}}^3\end{array}$

Correct Option is (B)

$\begin{array}{rl}W& =-P_{\text{ext }}\mathrm{\Delta }V\\ & =-2\mathrm{atm}\times (1.5\mathrm{L})\\ & =-3\mathrm{atm}\mathrm{L}\times 1.01325=-3.0398{\mathrm{dm}}^3\mathrm{bar}\end{array}$

Now, $1{\mathrm{dm}}^3\mathrm{bar}=100\mathrm{J}$

Hence, $-3.0398{\mathrm{dm}}^3$ bar $\times \frac{100\mathrm{J}}{1{\mathrm{dm}}^3\mathrm{bar}}=-303.98\mathrm{J}\cong -303.9\mathrm{J}$