Question

The vapor pressure of water at ${100^0}\,C$ is $760\,mm$ . What will be the vapor pressure at ${95^0}\,C$ ? The latent heat of water at this temperature range is $548\,cal\,{g^{ - 1}}$.
A. $674.3\,mm$
B. $434.3\,mm$
C. $634.3\,mm$
D. $638.6\,mm$

Answer 1

Hint: The Clausius equation relates the pressure and temperature with the molal heat of evaporation by the relation ${\log _{10}}\dfrac{{{p_2}}}{{{p_1}}} = \dfrac{{\Delta {\rm H}}}{{2.303R}}(\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}})$ where ${p_1},{p_2}$ are pressures at two different temperatures ${T_1},{T_2}$ and $\Delta H$ is the molal heat of evaporation.
The latent heat gives the heat of evaporation per gram of the substance whereas $\Delta H$ signifies the heat of evaporation per mole of the substance.

Formula used:
The Clausius equation is given as
${\log _{10}}\dfrac{{{p_2}}}{{{p_1}}} = \dfrac{{\Delta {\rm H}}}{{2.303R}}(\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}})$
Where ${p_1},{p_2}$ are pressures at two different temperatures ${T_1},{T_2}$ and $\Delta H$ is the molal heat of evaporation.
Conversions that will be useful are
$^0C\, + \,273\, = \,K$
$1J = 4.2\,cal$

Complete step by step solution:
We know that the Clausius equation is given as
${\log _{10}}\dfrac{{{p_2}}}{{{p_1}}} = \dfrac{{\Delta {\rm H}}}{{2.303R}}(\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}})$
According to the given question, let's take ${T_1} = {95^0}\,C$ and ${T_2} = {100^0}\,C$ .
Since we need to apply the Clausius equation we shall first convert the given temperatures to Kelvin units.
So, using the temperature scale conversion formula which is given as $^0C\, + \,273\, = \,K$ we will convert the given temperatures.
For ${T_1}$ ,
${T_1} = 95 + 273$
${T_1} = 368\,K$
For ${T_2}$ ,
${T_2} = 100 + 273$
${T_2} = 373\,K$
Accordingly, the pressures at the specific temperatures are given as ${p_2} = \,760\,mm$and we have to calculate${p_1}$
We are also given the latent heat of water in this temperature range which is $548\,cal\,{g^{ - 1}}$ .
Going by the basic definition, we can say that $548\,cal$ of heat will be released when $1\,g$of water evaporates. So, when $18\,g$ of water is vaporised, the amount of heat released known as molal heat of evaporation is given as
$(548 \times 18)\,cal$ .
Now we know that $1J = 4.2\,cal$
So, the molal heat of vaporization is given as $\Delta H = (\dfrac{{548 \times 18}}{{4.2}})\,J$
Solving this we get $\Delta H = 2348.57\,J$ .
Now substituting the values in the Clausius Equation
${\log _{10}}\dfrac{{{p_2}}}{{{p_1}}} = \dfrac{{\Delta {\rm H}}}{{2.303R}}(\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}})$
Where ${p_1} = \,?$ ,${p_2} = 760\,mm$ ,${T_1} = 368\,K$ ,${T_2} = 373\,K$ and $\Delta H = 2348.57\,J$ .
${\log _{10}}\dfrac{{760}}{{{p_1}}} = \dfrac{{2348.57}}{{2.303R}}(\dfrac{1}{{368}} - \dfrac{1}{{373}})$
Here R is the universal gas constant with the value $R = 8.314\,\,lit\,atm\,{K^{ - 1}}\,mo{l^{ - 1}}$ .
Hence the equation now becomes
${\log _{10}}\dfrac{{760}}{{{p_1}}} = \dfrac{{2348.57}}{{2.303 \times 8.314}}(\dfrac{1}{{368}} - \dfrac{1}{{373}})$
On further solving we get
${\log _{10}}\dfrac{{760}}{{{p_1}}} = \dfrac{{2348.57}}{{19.417}}(\dfrac{5}{{137264}})$
$ \Rightarrow {\log _{10}}\dfrac{{760}}{{{p_1}}} = 120.95 \times 0.000036$
$ \Rightarrow {\log _{10}}\dfrac{{760}}{{{p_1}}} = 0.0043$
Taking antilog both sides
$\dfrac{{760}}{{{p_1}}} = {10^{0.0043}}$
Solving this equation, we get
$\dfrac{{760}}{{{{10}^{0.0043}}}} = {p_1}$
$ \Rightarrow {p_1} = 634.3\,mm$
Hence option C is correct.

Note: The value of the universal gas constant, $R = 8.314\,\,lit\,atm\,{K^{ - 1}}\,mo{l^{ - 1}}$ is in Kelvin units but the units given in the question are not in the same units. So always first convert the temperatures in Kelvin, pressures in lit atm and the heat of evaporation in per moles.
Always follow the correct conversion rules to get the correct answer.