Question

10. What is the change in the boiling point of water at 1000 C per Pa change under atmospheric pressure conditions? The molar enthalpy of vaporization is 40.69 kJ mol-1, the molar volume of liquid water is 0.019 x 10-3 m3 mol-1, and the molar volume of steam is 30.199 x 10-3m3 mol-1, all at 1000 C and 1.01325 bar. (Hint: change in temperature per change in pressure)

Answer 1

The answer is
`(dT)/(dP) = 2.84*10^(-4)\ K/Pa`

Step-by-step explanation:

From the question we are told that

The boiling point of water is
`T_b = 100^oC = 100 + 273 = 373 \ K`

The enthalpy of vaporization is
`\Delta H_(vap) = 40.69 \ kJ mol-1 = 40.69 *10^(3) \ J/mol`

The molar volume of liquid water is
`V =0.019 * 10^(-3) m^3 mol^(-1)`

The molar volume of steam is
`V_s = 30.199 * 10^(-3) \cddot m^3\cdot mol^(-1)`

The pressure is
`P = 1.01325 \ bar`

Gnerally from Clausius Clapeyron equation we have that

`(dP)/(dT) = (\Delta H_(vap))/( T * \Delta V)`

Here
`\Delta V = V_s - V`

=>
`\Delta V = [30.199 * 10^(-3) ] - [ 0.019 * 10^(-3)]`

=>
`\Delta V = [30.199 * 10^(-3) ] - [ 0.019 * 10^(-3)]`

=>
`\Delta V = 0.03018 \ m^3 mol^(-1)`

So

`(dP)/(dT) = (40.69 *10^(3) )/( 373 * 0.03018)`

=>
`(dP)/(dT) = 3522.28 \ Pa/K`

Generally from the we are ask to obtain the change in the boiling point of water at 1000 C per Pa change under atmospheric pressure conditions which is mathematically represented as

`(d T)/(dP)`

So

`(dT)/(dP) = (1)/((dP)/(dT) ) = (1)/(3522.28)`

=>
`(dT)/(dP) = (1)/((dP)/(dT) ) = 2.84*10^(-4)\ K/Pa`