Kerala KEAM 5 June 2024 Solved Paper

The enthalpy change at constant pressure ($\Delta H_P$) and constant volume ($\Delta H_V$) are related by the equation: $$\Delta H_P = \Delta H_V + \Delta n \cdot R \cdot T$$where:- $\Delta H_P$ is the enthalpy change at constant pressure,- $\Delta H_V$ is the enthalpy change at constant volume,- $\Delta n$ is the change in the number of moles of gas between products and reactants,- $R$ is the universal gas constant (8.3 J K$^{-1}$ mol$^{-1}$),- $T$ is the temperature (300 K).Given:- The enthalpy of combustion at constant pressure, $\Delta H_P = -1410 \, kJ/mol$,- $R = 8.3 \, J/K mol$,- $T = 300 \, K$.For the combustion of ethylene, the change in the number of moles of gas is: $$\Delta n = 1 \quad (since ethylene combustion involves the same number of moles of gas molecules in reactants and products).$$Substitute the given values into the equation: $$\Delta H_V = \Delta H_P - \Delta n \cdot R \cdot T$$ $$\Delta H_V = -1410 - (1) \cdot (8.3) \cdot (300)$$ $$\Delta H_V = -1410 - 2490 = -1405 \, kJ/mol$$Thus, the enthalpy change at constant volume is $-1405 \, kJ/mol$, which corresponds to option (A). Quick Tip: For processes involving a change in pressure or volume, use the relation $\Delta H_P = \Delta H_V + \Delta n \cdot R \cdot T$ to find the enthalpy at constant volume. Ensure to account for the change in the number of moles of gas.