For benzene (C
_{6} H
_{6} ), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa?
Solution
If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, Δ
H_{vap,} then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation:
Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (
T_{1} = 80.1 °C = 353.3 K,
P_{1} = 101.3 kPa, Δ
H_{vap} = 30.8 kJ/mol) and want to find the temperature (
T_{2} ) that corresponds to vapor pressure
P_{2} = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for
T_{2} . Rearranging the Clausius-Clapeyron equation and solving for
T_{2} yields:
${T}_{2}={\left(\frac{-R\text{\u22c5}\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\frac{{P}_{2}}{{P}_{1}}\right)}{\text{\Delta}{H}_{\text{vap}}}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{{T}_{1}}\right)}^{\mathrm{-1}}={\left(\frac{-\left(8.3145\phantom{\rule{0.2em}{0ex}}\text{J/mol}\text{\u22c5}\text{K}\right)\text{\u22c5}\text{ln}\phantom{\rule{0.2em}{0ex}}\left(\frac{83.4\phantom{\rule{0.2em}{0ex}}\text{kPa}}{101.3\phantom{\rule{0.2em}{0ex}}\text{kPa}}\right)}{\text{30,800 J/mol}}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}\frac{1}{353.3\phantom{\rule{0.2em}{0ex}}\text{K}}\right)}^{\mathrm{-1}}=\text{346.9 K or}\phantom{\rule{0.2em}{0ex}}{73.8}^{\circ}\text{C}$
Check your learning
For acetone (CH
_{3} )
_{2} CO, the normal boiling point is 56.5 °C and the enthalpy of vaporization is 31.3 kJ/mol. What is the vapor pressure of acetone at 25.0 °C?
Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, Δ
H_{vap} . For example, the vaporization of water at standard temperature is represented by:
One way our body is cooled is by evaporation of the water in sweat (
[link] ). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at
T = 37 °C (normal body temperature); Δ
H_{vap} = 43.46 kJ/mol at 37 °C.
Solution
We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:
When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or
melting . At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (
[link] ).
Questions & Answers
how to know photocatalytic properties of tio2 nanoparticles...what to do now
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
In this morden time nanotechnology used in many field .
1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc
2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc
3- Atomobile -MEMS, Coating on car etc.
and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change .
maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
The eyes of some reptiles are sensitive to 850 nm light. If the minimum energy to trigger the receptor at this wavelength is 3.15 x 10-14 J, what is the minimum number of 850 nm photons that must hit the receptor in order for it to be triggered?
A teaspoon of the carbohydrate sucrose contains 16 calories, what is the mass of one teaspoo of sucrose if the average number of calories for carbohydrate is 4.1 calories/g?
4. On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of acetone (56.2 °C) and 1-propanol (97.4 °C), which have similar molar masses
Which of the following will increase the percent of HF that is converted to the fluoride ion in water? (a) addition of NaOH (b) addition of HCl (c) addition of NaF