Chemistry - Thermodynamics - I Question with Solution | TestHub

Question Explanation : Identify the mathematically incorrect expressions for the net change in enthalpy from state 1 to state 3

Concept: Definition of Enthalpy ( $H=U+PV$ ) and the First Law of Thermodynamics ( $\Delta U=q+w$ ). For an isochoric process, $\mathrm{w}=0$; for an adiabatic process, $\mathrm{q}=0$.

Solution:

We break down the thermodynamic changes for the two processes involving one mole of gas:

1. Process 1 → 2 (Heating at constant volume $V_1$):

The work done $w_1=0$. By the first law, $\Delta U_1=q_V=C_V\left(T_2-T_1\right)$.

2. Process 2 → 3 (Irreversible adiabatic compression against constant $P_3$):

No heat is exchanged, $q_2=0$. The energy change is due entirely to mechanical work done on the system:

$\Delta U_2=w_2=-P_{ext}(V_2-V_1)=-P_3(V_2-V_1)=-P_3{V}_1-P_3{V}_2$.

Net Enthalpy Change:

$$\begin{gathered} \text{displaylines}\Delta H=\Delta U_{total}+\Delta (PV{)}_{total} \\ \Delta U_{total}=\Delta U_1+\Delta U_2=C_V\left(T_2-T_1\right)+P_3{V}_1-P_3{V}_2 \\ \Delta (PV{)}_{total}=P_3{V}_2-P_1{V}_1 \end{gathered}$$.

Adding these terms:

$\Delta H_{}=C_V(T_2-T_1)+P_3{V}_1-P_3{V}_2)+\left[P_3{V}_2-P_1{V}_1\right]]$.

$\Delta H_{}=C_V(T_2-T_1)+P_3{V}_1-P_1{V}_1$.

Final Answer: Option (A), (B) and (D)