MHTCET 19 Apr 2026 Shift 2 Paper

Concept:For a gaseous reaction at constant temperature and pressure, the pressure-volume work is given by $w = -\Delta n_g RT$, where $\Delta n_g$ is the change in the number of moles of gaseous species.Explanation:The reaction is:$C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g)$Calculate the change in gaseous moles:$\Delta n_g = \text{(moles of gaseous product)} - \text{(moles of gaseous reactants)}$$\Delta n_g = 1 - (1 + 1) = -1$Convert temperature to kelvin:$T = 27 + 273 = 300 \text{ K}$Substitute into the work formula:$w = -\Delta n_g RT = -(-1)(8.314)(300)$$w = 2494.2 \text{ J}$Since $\Delta n_g$ is negative, the volume decreases, so work is done on the system and $w$ is positive.Answer:$w = 2494.2 \text{ J}$, hence Option A is correct.