Redding
Chemistry Homework Three
1. Three identical containers…
a. According to Avogadro’s Law, equal volumes of any two gases at
the same temperature and pressure contain the same number of
molecules. Because the flasks are all the same volume and are at
the same temperature and pressure, they must have the same
number of moles. The flasks are the same volume, and therefore
will contain the same number of molecules (particles) of any gas
present. It will be the same for the lighter gas, such is He, as for a
heavier gas such as NO2.
b. Flask B contains the largest total mass of gas. Although the flasks
contain the same number of particles, the mass of those particles
are different. NO2 molecules have a mass of approximately 46
g/mole, while CO has a mass of 28 g/mole and He has a mass of 4.
Both CO and He have masses that are less than NO2.
c. The gases will have the same average kinetic energy because it
depends only upon temperature of the gases. Since all three flasks
are at the same temperature, all three will have the same average
kinetic energy.
d. Although the gases have the same average kinetic energy, flask C
(He) will have the greatest average velocity because it has the
lightest (lowest) mass. Average velocity depends on temperature
as well as the mass of the gas particles. The temperature is the
same for all three flasks, so the mass is the only variable that will
influence the average velocity. (Colder gases behave like particles
with higher masses and will move slower then particles with a
higher temperature that behave like lighter gases)
2. Effusion
a. Determine the mole percent of each gas in the initial chamber
before effusion occurs.
He
1 mole He/ 5 moles total * 100 = 20% He
SO2
4 moles SO2 / 5 moles total * 100 = 80% SO2
b. Determine the ratio of moles He : moles SO2
Nx/ Ny * √My/Mx
1 mole He/ 4 moles SO2 * √64g/mol SO2 / 4 g/mol He
.25* √16=
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.25*4=1
The ratio is 1 mole of He : 1 mole of SO2 in chamber number 1.
1 mole He / 2 moles total * 100 = 50% He
1 mole SO2 / 2 moles total * 100 = 50 % SO2
The mole percent is of each gas in chamber 1 is now 50 % He and
50 % SO2.
c. Determine the ratio of moles of He : moles of SO2 in chamber
#2.
Nx/ Ny * √My/Mx
1 mole He/ 1 moles SO2 * √64g/mol SO2 / 4 g/mol He
1* √16=
1 *4=4
The ratio is 4 moles of He : 1 mole of SO2 in chamber number 2.
4 mole He / 5 moles total * 100 = 80% He
1 mole SO2 / 5 moles total * 100 = 20 % SO2
The mole percent is of each gas in chamber 1 is now 80 % He and
20 % SO2.
d. The effusion process creates a gaseous mixture that is
enriched in the gas of lower molar mass. Thus if more than
one effusion occurs, each newly effused gas is even more
enriched in the gas of lower molar mass.
The statement above is valid because the lighter gas, in this
case He, is able to move through the pinhole faster than the larger
gas which in this case is the SO2. When the gases effused from the
initial chamber to chamber one, the ratio of He to SO2 was 1:1.
Although at first glance a 1:1 ratio does not seem to indicate that
the mixture is more enriched with He, it actually is. The initial
amounts of gases also need to be taken into consideration. The
initial amount of He was 1 mole whereas the initial amount of SO2
was 4 moles, therefore more of the He had to effuse through the
pin hole to achieve a 1:1 ratio. Because there were now equal
amounts of He and SO2, when the gases effused from chamber
one to chamber two, the effused gas was certainly more enriched
with the He. The ratio was now 4 moles of He to 1 mole of SO 2,
meaning there was now 80% He and only 20 % SO2.
3. “Real” Gases
Redding
a. How would the pressure (P) be altered from its idea gas value
(Pideal) due only to modified postulate #1?
P>Pideal
Modified postulate 1 states that gas particles have volume. There is
less space available due to the particle size, so the available area of the
container decreases. The volume (free space) decreases due to the
volume that the spheres take up. If P=f/a by definition, and the area
decreases due to the volume of the particles occupying space, then the P
would have to be greater than the Pideal. If the area decreases, the
pressure must increase.
b. How would the pressure (P) be altered from its idea gas value
(Pideal) due only to modified postulate #2?
P<Pideal
In this case, the particles have no volume but now we will assume
the particles have attractive forces between each other. I would
imagine that a large percentage of the molecules in the container are
surrounded evenly by other molecules and will not feel the effects of
the intermolecular forces due to balanced forces. However, not all of
the molecules are surrounded by other identical particles. Some of the
particles may be next to a wall and experience an unbalanced force in
which they may be pulled towards the particles and away from the wall.
This may result in particles hitting the wall with less force or perhaps
not at all. This lower force would only occur if there were an attraction
between particles. Again stating that the definition of pressure is P=f/a,
then it may be stated that if the force decreases then the pressure
decreases.
c. i. According to the ideal gas law, Pideal for each gas should be the
same. Why?
According the ideal gas law, the pressure should be the same
because the particles are assumed to take up no space and have no
intermolecular forces between them. If that is the case, then their
pressure should be the same at the same volume and temperature.
The force and the area would be the same, so the pressure would
therefore be the same.
c. ii. Considering each gas separately, which “non-ideal” factor is
contributing most to the measured pressure?
(ideal pressure =12atm)
In gas X, the molecular volume is contributing more to the
pressure because the pressure is greater than 12 atm. In order for
the pressure to be greater, the volume must increase (as stated in
3a). In gas Y, the attractive forces are contributing more to the
Redding
pressure because the pressure is less than 12 atm. In order for the
pressure to decrease, the attractive forces should decrease (as
stated in 3b).
4. Specific Heat
Lead: 0.128 J g-1Co-1
Calculate the value in cal g-1Co-1 , cal g-1Ko-1, and cal mol-1Co-1
cal g-1Co-1
4.184J = 1 cal
(0.128 J g-1Co-1)(1 cal g-1Co-1/ 4.184J) = 0.031 cal g-1Co-1
cal g-1Ko-1
Because 1 degree on the C scale is equal to 1 degree on the K scale
(discussed in class) specific heat can be reported in either cal g-1Co-1
or cal g-1Ko-1 . The answer would be 0.031 cal g-1Ko-1. This would NOT
be the case for Fo!
cal mol-1Co-1
(0.031 cal g-1Co-1)(207.2g/mol) = 6.4 cal mol-1Co-1
Extra Credit:
Discuss what his experiments had to do with heat flow, temperature
change and waterfalls. What was the purpose of the experiment?
What was Joule trying to prove?
Joule hypothesized that when water fell approximately 800 feet, it
would rise one degree Fahrenheit. Many of the sources that I researched
had conflicting accounts of what actually happened on his now famous
honeymoon. One source states that while on his honeymoon in Chamonix
in the Alps, Joule found that the temperature at the base of a waterfall was
higher than the temperature at the top of the waterfall. He concluded that
this meant that the energy of the falling water was being converted to
heat. This was a major step in proving the theory of energy conservation,
which is known as the First Law of Thermodynamics
(http://www.ulearntoday.com/magazine/physics_article1.jsp?FILE=joule). Another source states
that although he did attempt this experiment, he was unsuccessful due to
the height of the waterfall and the spray produced at the bottom of the
waterfall http://www.zephyrus.co.uk/jamesjoule.html) There was also a rumor that he
brought a huge thermometer with him, but it was thought to be made up
by Lord Kelvin (http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/Joule.htm)
In conclusion, Joule was trying to prove the law of conservation of
energy. He thought that the kinetic energy of the water was converted to
heat, and therefore the temperature of the water at the bottom of a
waterfall would be one degree Fahrenheit higher than the temperature at
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the top of the waterfall. The waterfall had to be of a certain height, as
stated previously, approximately 800 feet or 300m.