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

Answer:

The answer  is  $\frac{dT}{dP} = 2.84*10^{-4}\ K/Pa$

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

      $\frac{dP}{dT} = \frac{\Delta H_{vap}}{ T * \Delta V}$

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

    $\frac{dP}{dT} = \frac{40.69 *10^{3} }{ 373 * 0.03018}$

=>  $\frac{dP}{dT} = 3522.28 \ Pa/K$

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

       $\frac{d T}{dP}$

So

    $\frac{dT}{dP} = \frac{1}{\frac{dP}{dT} } = \frac{1}{3522.28}$

=> $\frac{dT}{dP} = \frac{1}{\frac{dP}{dT} } = 2.84*10^{-4}\ K/Pa$