If water vapour is assumed to be a perfect gas, molar enthalpy change

1.	If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporization of 1 mol of water at 1 bar and 100°C is 41 kJ mol-1 Calculate the internal energy change, when: (i) 1 mol of water is vaporized at 1 bar pressure and 100°C. (ii) 1 mol of water is converted into ice.
Answer» (i) H₂O(l) → H₂O(g) Δn_(g)= 1 − 0 = 1 \(\because\) ΔH = ΔU + Δn_gRT or ∆U = ∆H − ∆n_gRT Given, ∆H = 41 kJ mol⁻¹, T = 100 + 273 = 373 K \(\therefore\) ∆U = 41 kJ mol⁻¹− 1 × 8.3 J mol⁻¹K⁻¹ × 373 K = 37.904 kJ mol⁻¹ (ii) H₂O(l) → H₂O(s) There is negligible change in volume. \(\therefore\) p∆V = ∆n_gRT = 0 \(\therefore\) ∆H = ∆U Or ∆U = 41 kJ mol⁻¹

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

Answer»

(i) H₂O(l) → H₂O(g)

Δn_(g)= 1 − 0 = 1

\(\because\) ΔH = ΔU + Δn_gRT

or ∆U = ∆H − ∆n_gRT

Given, ∆H = 41 kJ mol⁻¹,

T = 100 + 273 = 373 K

\(\therefore\) ∆U = 41 kJ mol⁻¹− 1 × 8.3 J mol⁻¹K⁻¹ × 373 K

= 37.904 kJ mol⁻¹

(ii) H₂O(l) → H₂O(s)

There is negligible change in volume.

\(\therefore\) p∆V = ∆n_gRT = 0

\(\therefore\) ∆H = ∆U

Or ∆U = 41 kJ mol⁻¹

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

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