If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water - Chemistry - Thermodynamics

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

Priya G N · Accepted Answer

Molar enthalpy equation for an ideal or perfect gas is

is the molar enthalpy = 41kJ mol -1

is the internal energy change - to be calculated

is the no of moles created or absorbed during the chemical reaction R is the gas constant = 8 314 J mol -1 K-1 T is temperature in K = 100 degree C = 373 K The internal energy change, when (i) 1 mol of water is vaporised at 1 bar pressure and 100ï¿½C Let us consider the reaction,

=

= 3-2 = 1 Substituting in molar enthalpy equation for an ideal or perfect gas, the imternal energy change is,

The internal energy change, when (ii) 1 mol of water is converted into ice Let us consider the reaction,

=

= 1-1 = 0 Substituting in molar enthalpy equation for an ideal or perfect gas, the imternal energy change is,

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.

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.