Question

For water \(\Delta_{vap}H=41\) kJ mol\(^{-1}\) at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is __________ kJ mol\(^{-1}\). (Nearest integer)
[Use: R=8.3 J mol\(^{-1}\)K\(^{-1}\)]

Hint:

When using the formula \(\Delta H = \Delta U + \Delta n_g RT\), pay close attention to units. \(\Delta H\) is often given in kJ, while R is in J. You must convert them to the same unit before adding or subtracting. A common mistake is forgetting to convert kJ to J (or vice versa), leading to a significantly wrong answer.

Answer 1

Correct Answer: 38

To find the internal energy change (\(\Delta U\)) during the evaporation of water, we use the relationship between enthalpy change (\(\Delta H\)) and internal energy change. The formula is:

\(\Delta H = \Delta U + \Delta nRT\)

Here, \(\Delta H\) is 41 kJ mol\(^{-1}\) (given), \(R = 8.3\) J mol\(^{-1}\)K\(^{-1}\), and \(T = 373\) K.

Since water vapor is treated as an ideal gas, evaporation involves converting liquid water to water vapor, increasing the moles of water from 1 mole of liquid to 1 mole of vapor. The change in moles of gas (\(\Delta n\)) is thus 1 mol (since 1 mole of liquid water produces 1 mole of gaseous water).

Convert \(R\) from J to kJ:

\(R = \frac{8.3 \text{ J mol}^{-1}\text{K}^{-1}}{1000} = 0.0083 \text{ kJ mol}^{-1}\text{K}^{-1}\)

Now calculate \(\Delta nRT\):

\(\Delta nRT = 1 \times 0.0083 \times 373 = 3.0959 \text{ kJ mol}^{-1}\)

Substituting values into the enthalpy formula:

\(\Delta U = \Delta H - \Delta nRT = 41 - 3.0959 = 37.9041 \text{ kJ mol}^{-1}\)

Rounded to the nearest integer, the internal energy change (\(\Delta U\)) is 38 kJ mol\(^{-1}\).

Confirming the solution, 38 kJ mol\(^{-1}\) falls within the expected range of 38-38.

Therefore, the internal energy change during evaporation is 38 kJ mol\(^{-1}\).