Chemistry - Thermodynamics - I Question with Solution | TestHub

Since pressure is constant, it is an irreversible process.

$$\begin{array}{rlrl}\Rightarrow \mathrm{W}& =-\mathrm{P}\left({\mathrm{V}}_2-{\mathrm{V}}_1\right)& =-1(20-10)& =-10\mathrm{atm}\mathrm{L}=-1013\mathrm{J}\end{array}$$

Hence, choice (A) is correct

Being an adiabatic expansion, temperature is not constant. Hence,

$$\begin{array}{rlr}& \Delta \mathrm{U}=\mathrm{q}+\mathrm{W}& \Delta \mathrm{U}=\mathrm{W}=-1013\mathrm{J}\end{array}$$

Hence, choice (B) is correct.

$$\begin{array}{r}\Delta H=n{C}_p\Delta T;\Delta U=n{C}_v\Delta T\Rightarrow \frac{\Delta \mathrm{H}}{\Delta \mathrm{U}}=\frac{{\mathrm{C}}_{\mathrm{P}}}{{\mathrm{C}}_{\mathrm{V}}}\end{array}$$

As the gas is monoatomic, ${\mathrm{C}}_{\mathrm{P}}=\frac{5}{2}\mathrm{R}$ and ${\mathrm{C}}_{\mathrm{V}}=\frac{3}{2}\mathrm{R}$

$$\frac{\frac{5}{2}R}{\frac{3}{2}R}=\frac{5}{3}$$

$\Rightarrow \Delta \mathrm{H}=\frac{5}{3}\Delta \mathrm{U}=-1688.33\mathrm{J}$

Hence, choice (C) is correct.

$$\Delta \mathrm{U}={\mathrm{nC}}_{\mathrm{v}}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=\frac{3}{2}\times \mathrm{R}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right)=-1013\mathrm{J}$$

$\Rightarrow {\mathrm{T}}_2=218.96\mathrm{K}$