Question

A fish swimming in water body when taken out from the water body is covered with a film of water of weight 36 g. When it is subjected to cooking at 100 °C, then the internal energy for vaporization in kJ mol^–1 is ________. [nearest integer]
[Assume steam to be an ideal gas. Given Δ_vapH^Θ for water at 373 K and 1 bar is 41.1 kJ mol^–1; R = 8.31 J K^–1 mol^–1]

Answer 1

Correct Answer: 38

To determine the internal energy for vaporization (\(\Delta_{vap}U^{\Theta}\)) at 100 °C, we use the relationship between enthalpy and internal energy: \(\Delta_{vap}H^{\Theta} = \Delta_{vap}U^{\Theta} + nRT\). Here, \(n = 1\) mol, \(R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1} = 0.00831 \text{ kJ K}^{-1} \text{ mol}^{-1}\), and \(T = 373 \text{ K}\). Rearrange the equation for \(\Delta_{vap}U^{\Theta}\): \(\Delta_{vap}U^{\Theta} = \Delta_{vap}H^{\Theta} - nRT\). Substitute the values: \(\Delta_{vap}U^{\Theta} = 41.1 \text{ kJ mol}^{-1} - 1 \times 0.00831 \text{ kJ K}^{-1} \times 373 \text{ K}\). Calculate the value: \(\Delta_{vap}U^{\Theta} = 41.1 \text{ kJ mol}^{-1} - 3.1 \text{ kJ mol}^{-1} = 38.0 \text{ kJ mol}^{-1}\). Verify if this value is within the range [38,38]; it is exact. Thus, the internal energy for vaporization is 38 kJ mol^-1.