Question

If water vapour is assumed to be a perfect

gas, molar enthalpy change for

vapourisation of 1 mol of water at 1bar

and 100°C is 41kJ mol–1. Calculate the

internal energy change, when

(i) 1 mol of water is vaporised at 1 bar

pressure and 100°C.

(ii) 1 mol of water is converted into ice.​

Answer 1

$\huge\underline\frak{\fbox{AnSwEr :-}}$

Solution

(i) The change H O H O g 2 2 (l ) → ( )

∆ = ∆ + ∆ H U n T gR

or ∆U = ∆H – ∆n T gR , substituting the

values, we get

41.00 kJ mol 1

8.3 J mol K 373 K

= 37.904 kJ mol–1

(ii) The change H O H O s 2 2 (l)

There is negligible change in volume,

So, we can put p V∆ = ≈ ∆n T gR 0 in this

case,

∆ ≅ ∆ H U1

so, 41.00kJ mol

Answer 2

Answer:

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