Question

If the normal boiling point of acetone is ${56^ \circ }C\;$ and it has a $\Delta {H_{vap}}\;$ of $32.1{\text{ }}kJmo{l^{ - 1}}$ , estimate the boiling point at $5$ bar?

Answer 1

The previous problem is solved with the Clausius Clapeyron equation. First we know the equation and then we solve the problem. The Clausius Clapeyron equation gives the relationship between pressure and temperature under two-phase equilibrium conditions. The two phases can be vapor and solid, forming solid and liquid sublimation to melt.

Complete Step By Step Answer:
Enthalpy of vaporization, also known as (latent) heat of vaporization or heat of vaporization, is the energy (enthalpy) that must be added to a liquid substance to convert a certain amount of the substance into a gas.
The normal boiling point of acetone is measured at $1$ bar.
This type of problem and solved by using clausius clapeyron equation which is given as follows:
$\ln \left( {\dfrac{{{P_1}}}{{{P_2}}}} \right) = \dfrac{{\Delta {H_{vap}}}}{R}\left( {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right)$
$\Delta {H_{vap}}\; = 32.1{\text{ }}kJmo{l^{ - 1}}$
$R$ is the universal gas constant.
$R = {\text{0}}{\text{.008134 KJ(molK}}{{\text{)}}^{{\text{ - 1}}}}$
${P_1} = 1bar$
${P_2} = 5bar$
${T_1} = 56 + 273.15$
${T_1} = 329.15K$
We have to find out ${T_2}.$
Now, we will plug in the given values in the formula above to find the boiling point of acetone.
$\ln \left( {\dfrac{1}{5}} \right) = \dfrac{{32.1}}{{0.008134}}\left( {\dfrac{1}{{329.15}} - \dfrac{1}{{{T_2}}}} \right)$
$- 1.6094 = 3946.3978\left( {\dfrac{1}{{329.15}} - \dfrac{1}{{{T_2}}}} \right)$
On further solving the above equation, we get
${T_2} = \;381.5{\text{ }}K$
Therefore, the boiling point of acetone at five bar pressure is $381.5{\text{ }}K$ .
We can convert this value to Celsius scale as follows:
${T_2} = 381.5{\text{ - 273}}{\text{.15}}$
${T_2} = {108^ \circ }C$
Hence, the boiling point of acetone at five bar pressure is ${108^o}C$ .

Note:
We should remember that the relation between the temperature in Celsius scale and kelvin scale is given as follows: ${T^o}C = T(K) - 273.15$ . To solve these types of questions where initial and final pressures are given along with the initial temperature, we simply use the clausius clapeyron equation.