Clausius-Clapeyron Equation, and Cubic Unit Cells

In summary, the normal boiling point of acetone is 56.5 oC and its heat of vaporization is 32.0 kJ/mol. The boiling point at 580 mmHg can be found using the formula \frac{lnP_2 - lnP_1}{\frac{1}{T_2} - \frac {1}{T_1}} = - \frac{\Delta H}{R}. For question 2, the atomic radius of an element in body centered cubic structure with a density of 7.19 g/mL can be found by determining the mass of a single unit cell, converting it to amu, and using the formula \frac {\sqrt{3}s}{4} =
  • #1
Silverious
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1. Acetone has normal boiling point of 56.5 oC, and heat of vaporization of 32.0 kJ/mol. Find the boiling point at 580 mmHg.


2. Given element, its structure and density, find its atomic radius in pm. Cr, body centered cubic, 7.19 g/mL.


Well, as for question 1, I'm totally stumped. Everytime I try to go about doing any problem like this, my brain locks up. I think I just don't have a firm understanding of the idea.

Number 2, well, umm...again, same as above. I really just don't understand anything. It's pretty frustrating. So if someone could help me pleaseeeee, I will become their slave.
 
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  • #2
I've thought about the first question a bit more.

Is [tex] \frac{lnP_2 - lnP_1}{\frac{1}{T_2} - \frac {1}{T_1}} = - \frac{\Delta H}{R} [/tex] what I'm looking for?
 
  • #3
I'm not crazy, talking to myself am I?

Anyways, regarding the second question.

First I have to determine the mass of a single unit cell.

I believe, this is found by determining the molar mass of Cr, then converting it to amu's. Since there are 2 nuclei in a body centered cubic cell, it would be 2x Molar Mass(in amu) correct?

Now, in order to determine the actual radius, I need to use density fomrula. [tex] V_{unit cell} = \frac {m_{unit cell}}{d} [/tex]

But first I convert mL into cubic whats? I forget...oh well. Anyways, from there, I convert to picometers, then I take the cube of this number, which is "s". But I can't use this yet. Sooooo, I use [tex] \frac {\sqrt{3}s}{4} = r [/tex].


Someone please verify these?
 
  • #4
Is no one going to reply to this? Or should I post these in chemistry forum?
 
  • #5
You're doing fine --- you got the concepts --- at that point, it becomes an exercise in arithmetic.
 

1. What is the Clausius-Clapeyron Equation?

The Clausius-Clapeyron Equation is a mathematical relationship that describes the behavior of the vapor pressure of a substance as it undergoes a phase change between two phases, usually between a solid and a liquid or between a liquid and a gas.

2. How is the Clausius-Clapeyron Equation derived?

The Clausius-Clapeyron Equation is derived from the ideal gas law and the principle of phase equilibrium. It can also be derived using statistical mechanics and thermodynamics principles.

3. What is the importance of the Clausius-Clapeyron Equation?

The Clausius-Clapeyron Equation is a fundamental tool in understanding phase transitions and predicting the behavior of substances under different temperature and pressure conditions. It is used in various industries, such as in refrigeration and in the production of pharmaceuticals.

4. What are cubic unit cells?

Cubic unit cells are the simplest and most symmetrical unit cells in crystal structures. They have equal sides and angles, and their atoms are arranged in a repeating pattern along the three axes. There are three types of cubic unit cells: simple cubic, body-centered cubic, and face-centered cubic.

5. How are cubic unit cells used in materials science?

Cubic unit cells are used in materials science to understand the physical and chemical properties of materials. They are also used in the design and synthesis of new materials, as well as in the analysis of crystal structures and defects in materials.

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