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Problems · Problem 5.7

Q.A swimmer coming out from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporisation at 298 K. ΔvapH⊖\Delta_{vap}H^\ominus for water at 298 K = 44.01 kJ mol−1^{-1}.

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✓ Free question

The heat needed to evaporate 18 g of water at 298 K is 44.01 kJ, and the internal energy of vaporisation at this temperature is 41.53 kJ mol⁻¹ — the difference arises because expansion work is done against the atmosphere.

The key insight here is that the enthalpy of vaporisation (ΔvapH⊖\Delta_{vap}H^\ominus) already includes the work done by the system when water vapour pushes against the atmosphere. But internal energy (ΔvapU⊖\Delta_{vap}U^\ominus) is the true energy change within the water molecules themselves — the energy needed to overcome intermolecular forces without counting the work done on the surroundings.

When you have 18 g of water, that's exactly 1 mole (since the molar mass of water is 18 g mol⁻¹). So the heat required at constant pressure is simply the molar enthalpy of vaporisation.


  1. Heat required to evaporate the water

    The mass of water is 18 g, which is exactly 1 mole. At constant pressure (the poolside is open to the atmosphere), the heat supplied equals the enthalpy change:

qp=ΔvapH⊖×n=44.01 kJ mol−1×1 mol=44.01 kJq_p = \Delta_{vap}H^\ominus \times n = 44.01\ \text{kJ mol}^{-1} \times 1\ \text{mol} = 44.01\ \text{kJ}

So 44.01 kJ must be supplied to evaporate the film.

  1. Relating enthalpy and internal energy

    For any process at constant pressure, the enthalpy change and internal energy change are related by:

ΔH=ΔU+Δ(PV)\Delta H = \Delta U + \Delta (PV)

For a phase change involving a gas, the volume of the liquid is negligible compared to the vapour. Assuming water vapour behaves as an ideal gas:

Δ(PV)=PΔV≈PVgas=nRT\Delta (PV) = P\Delta V \approx PV_{gas} = nRT

Therefore:

ΔvapH⊖=ΔvapU⊖+nRT\Delta_{vap}H^\ominus = \Delta_{vap}U^\ominus + nRT

  1. Calculate the internal energy of vaporisation

    Rearranging:

ΔvapU⊖=ΔvapH⊖−nRT\Delta_{vap}U^\ominus = \Delta_{vap}H^\ominus - nRT

For 1 mole at 298 K:

nRT=1×8.314 J K−1mol−1×298 K=2478 J mol−1=2.478 kJ mol−1nRT = 1 \times 8.314\ \text{J K}^{-1}\text{mol}^{-1} \times 298\ \text{K} = 2478\ \text{J mol}^{-1} = 2.478\ \text{kJ mol}^{-1}

So:

ΔvapU⊖=44.01−2.478=41.53 kJ mol−1\Delta_{vap}U^\ominus = 44.01 - 2.478 = 41.53\ \text{kJ mol}^{-1}

Watch out

A common mistake is to forget that RR must be in kJ when subtracting from ΔH\Delta H given in kJ. Using R=8.314 J K−1mol−1R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} gives an answer in joules — always convert to kJ before subtracting.

Tip

The term nRTnRT (about 2.48 kJ at 298 K) represents the work done by the vapour as it expands against atmospheric pressure. This is why ΔU\Delta U is smaller than ΔH\Delta H — some of the heat supplied goes into doing work, not into raising the internal energy.

✓Final answer

The heat required is 44.01 kJ and the internal energy of vaporisation at 298 K is 41.53 kJ mol⁻¹.

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