Question

004. If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol
of water at 1 bar and 100°C is 41 kJ mol-1. Calculate the internal energy change, when
a) 1 mol of water is vaporized at 1 bar pressure and 100°C.
b) 1 mol of water is converted into ice.​

Answer 1

Hi,

Answer:

The internal energy changes in the two processes are as follows:

(i) ΔU = 37.904 kJ/mol

(ii) ΔU = 41 kJ mol-1

Explanation:

Given data:

No. of moles of water, n = 1 mol

Pressure, P = 1 bar

Temperature, T = 100℃ = 100 + 273 K = 373 K

The molar enthalpy change for vapourisation, ΔH = 41 kJ mol-1

Case (i): when 1 mol of water is vaporized at 1 bar pressure and 100°C.

The change: H2O (l) → H2O(g)

We have, the molar enthalpy equation for an ideal gas as,  

ΔH = ΔU + ΔnRT  

⇒ ΔU = ΔH – ΔnRT  

⇒ ΔU = 41.00 – [1 * 8.3 * 373 * 10⁻³] …. [Substituting the given values & R = 8.3 J/K/mol]

⇒ ΔU = [41 kJ mol−1] – [3.096 kJ mol−1]

⇒ ΔU = 37.904 kJ/mol

Case (ii):  when 1 mol of water is converted into ice

The change: H2O (l) → H2O(s)

Since during this process, there will be a negligible change in the volume, so,  we can put

pΔV = ΔnRT ≈ 0.

So, in this case, ΔH = ΔU

∴ ΔU = 41 kJ mol−1

Hope this is helpful!!!!