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NEET CHEMISTRYThermodynamicsMedium

Question

The standard enthalpy of vaporisation ΔvapH\Delta_{vap}H^{\circ} for water at 100C100^{\circ}\text{C} is 40.66 kJ mol140.66 \text{ kJ mol}^{-1}. The internal energy of vaporisation of water at 100C100^{\circ}\text{C} (in kJ mol1\text{kJ mol}^{-1}) is: (Assume water vapour to behave like an ideal gas)

A

+37.56

B

-43.76

C

+43.76

D

+40.66

Step-by-Step Solution

For the vaporisation of water: H2O(l)H2O(g)H_2O(l) \rightarrow H_2O(g) The change in the number of moles of gaseous species, Δng=np(g)nr(g)=10=1\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - 0 = 1. The relationship between enthalpy change (ΔH\Delta H) and internal energy change (ΔU\Delta U) is given by : ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT Rearranging the formula to find ΔU\Delta U: ΔU=ΔHΔngRT\Delta U = \Delta H - \Delta n_g RT Given data: ΔH=40.66 kJ mol1\Delta H = 40.66 \text{ kJ mol}^{-1} T=100C=373 KT = 100^{\circ}\text{C} = 373 \text{ K} R=8.314 J K1mol1=8.314×103 kJ K1mol1R = 8.314 \text{ J K}^{-1} \text{mol}^{-1} = 8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{mol}^{-1} Substituting the values: ΔU=40.66 kJ mol1(1×8.314×103 kJ K1mol1×373 K)\Delta U = 40.66 \text{ kJ mol}^{-1} - (1 \times 8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{mol}^{-1} \times 373 \text{ K}) ΔU=40.663.101 kJ mol1\Delta U = 40.66 - 3.101 \text{ kJ mol}^{-1} ΔU=+37.559 kJ mol1+37.56 kJ mol1\Delta U = +37.559 \text{ kJ mol}^{-1} \approx +37.56 \text{ kJ mol}^{-1}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsstandardenthalpyvaporisationdeltavaphcirccirctextc

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