Chemistry - Thermodynamics - II Question with Solution | TestHub

(A) for process CA

$$\begin{array}{rlrl}& \mathrm{H}=\mathrm{U}+\mathrm{PV}& \Delta \mathrm{H}=\Delta \mathrm{U}+{\mathrm{P}}_{\mathrm{F}}{\mathrm{V}}_{\mathrm{F}}-{\mathrm{P}}_{\mathrm{I}}{\mathrm{V}}_{\mathrm{I}}& \end{array}$$

$$\Delta {\mathrm{H}}_{\mathrm{CA}}=\Delta {\mathrm{U}}_{\mathrm{CA}}+{\mathrm{P}}_1{\mathrm{V}}_1-{\mathrm{P}}_3{\mathrm{V}}_2$$

And

$$\Delta {\mathrm{H}}_{\mathrm{CA}}=\Delta {\mathrm{U}}_{\mathrm{CA}}+{\mathrm{nRT}}_1-{\mathrm{nRT}}_3$$

Since ${\mathrm{T}}_1>{\mathrm{T}}_3$

$$\begin{array}{rlrlrl}& \Delta {\mathrm{H}}_{\mathrm{CA}}=\Delta {\mathrm{U}}_{\mathrm{CA}}+\mathrm{nR}\left({\mathrm{T}}_1-{\mathrm{T}}_3\right)& \Delta {\mathrm{H}}_{\mathrm{CA}}>\Delta {\mathrm{U}}_{\mathrm{CA}}& & & \end{array}$$

$$\left(\Delta {\mathrm{U}}_{\mathrm{CA}}=\Delta {\mathrm{U}}_{\mathrm{CB}}+\Delta {\mathrm{U}}_{\mathrm{BA}}\right)$$

$\left(\Delta {\mathrm{U}}_{\mathrm{CA}}=\Delta {\mathrm{U}}_{\mathrm{CB}}\right)$

$$\Delta {\mathrm{H}}_{\mathrm{CA}}>\Delta {\mathrm{U}}_{\mathrm{CB}}(correct)$$

(B) for process CA

$$\Delta {\mathrm{U}}_{\mathrm{CA}}={\mathrm{q}}_{\mathrm{CA}}+{\mathrm{W}}_{\mathrm{CA}}$$

Since there is compression

$$\begin{array}{rlr}& {\mathrm{W}}_{\mathrm{CA}}>0& \Delta {\mathrm{U}}_{\mathrm{CA}}>{\mathrm{q}}_{\mathrm{CA}}\end{array}$$

And $\Delta {\mathrm{H}}_{\mathrm{CA}}>\Delta {\mathrm{U}}_{\mathrm{CA}}$ (correct)

(C) $\Delta {\mathrm{H}}_{\mathrm{BC}}={\mathrm{nC}}_{\mathrm{P}}\Delta \mathrm{T}$

$$\begin{array}{rlrl}& \Delta {\mathrm{U}}_{\mathrm{BC}}={\mathrm{nC}}_{\mathrm{v}}\Delta \mathrm{T}& \text{ Since }\Delta \mathrm{T}=-\mathrm{ve}& \Delta {\mathrm{H}}_{\mathrm{BC}}<\Delta {\mathrm{U}}_{\mathrm{BC}}\text{ (correct) }\end{array}$$

(D) $\Delta {\mathrm{H}}_{\mathrm{AB}}=\Delta {\mathrm{H}}_{\mathrm{BC}}+\Delta {\mathrm{H}}_{\mathrm{CA}}=0$

$$\begin{array}{rlrl}& \Delta {\mathrm{U}}_{\mathrm{AB}}={\mathrm{q}}_{\mathrm{AB}}+{\mathrm{W}}_{\mathrm{AB}}=0& {\mathrm{q}}_{\mathrm{AB}}=-{\mathrm{W}}_{\mathrm{AB}}& {\mathrm{q}}_{\mathrm{AB}}=\left|{\mathrm{W}}_{\mathrm{AB}}\right|\text{ (incorrect) }\end{array}$$