Transcript Gas Laws - University of California, Irvine

Gas Laws
Advanced Problem Solving
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SURVEY QUESTIONS
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• How to know what specific formulas to use for specific gas law problems
• Use PV=nRT if only one set of conditions are given, in which case nothing cancels.
• Use ratio form, if two sets of conditions are given. Cancel anything that is constant.
𝑃1𝑉1 𝑃2𝑉2
=
𝑛1𝑇1 𝑛2𝑇2
• Will we has problems that use limiting reagents and ideal gas laws? Will the final contain ideal gas law
problesm?
• Absolutely and yes. The final will contain everything.
• A myriad of questions about conversions between moles and grams, mol ratios, and using chemical reactions for
mol ratios.
• Please come see me/ Paul/ Tutors in office hours.
• Do we need to know {insert whatever conversion you are wondering about here}?
• No, they are on the front page of the exam equation sheet that you already have.
DALTON’S LAW
• Dinitrogen oxide gas was generated from the thermal decomposition of
ammonium nitrate and collected over water. The wet gas occupied 126 mL
at 21 C when the atmospheric pressure was 755 Torr. What volume would
the same amount of dry Dinitrogen oxide have occupied if collected at 755
Torr and 21 C. The vapor pressure of water is 18.65 Torr at 21 C.
Pressure of N2O:
Atmospheric =P(N2O)+P(H20vapor)
Hint 1: If the total pressure is 755 Torr,
what is the pressure of N2O?
755-18.65 torr=736.35 torr
Hint 2: How do you determine the
volume of a gas at a new pressure?
Then use:
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736.35 ∗ 126
𝑉2 =
755
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V2=123mL
REACTIONS: SINGLE GAS PRODUCT
• Suppose that 200.0 mL of propane gas C3H8 at 1.00 atm and 298K is mixed with 1.00 L of
oxygen gas at the same pressure and temperature and burned to form carbon dioxide gas and
liquid water. Determine the final volume of the reaction mixture at 2.00 atm and 298 K if the
reaction goes to completion. (note: volume of liquid water is negligible).
Hint1: Write out the
equation and balance it.
Balance Carbons and
Hydrogens first then
oxygens.
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1C3H8:5O2
mol ∝ V
mol ratio= volume ratio
(At constant temp/pressure)
REACTIONS:
Hint 2: Check for any limiting
reagents, which may leave
some gaseous reactant
unreacted.
You could solve for moles of
CO2 here, but that would be
a lot of calculations and a
lot of time/possibilities for
mistakes. Instead look for
how much volume would be
made at the SAME temp and
pressure, and then alter it for
the new temperature and
pressure.
Short way: more thinking, more logic jumps, less math
• Suppose that 200.0 mL of propane gas C3H8 at 1.00 atm and 298K is mixed with
1.00 L of oxygen gas at the same pressure and temperature and burned to form
carbon dioxide gas and liquid water. Determine the final volume of the reaction
mixture at 2.00 atm and 298 K if the reaction goes to completion. (note: volume of
liquid water is negligible).
1C3H8 : 5O2
mol ∝ V
mol ratio= volume ratio
(At constant temp/pressure)
200.0mL of propane requires
1.00L of O2
We have exactly that, no
excess reagent, reaction goes
to C3H8=O2=0atm
You may fill in and solve
OR
Pressure doubles, so
volume…….
Halves
Volume=300.mL
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Long way: less thinking, less logic jumps, more math
REACTIONS: SINGLE GAS PRODUCT
• Suppose that 200.0 mL of propane gas C3H8 at 1.00 atm and 298K is mixed with 1.00 L of oxygen gas at the same
pressure and temperature and burned to form carbon dioxide gas and liquid water. Determine the final volume of the
reaction mixture at 2.00 atm and 298 K if the reaction goes to completion. (note: volume of liquid water is negligible).
𝑛𝐶3𝐻8
𝑛𝑂2
𝑃𝑉
1.00𝑎𝑡𝑚 ∗ 0.2000𝐿
3𝑚𝑜𝑙 𝐶𝑂2
=
=
= 0.008175 𝑚𝑜𝑙 𝐶3𝐻8 ∗
= 0.245 𝑚𝑜𝑙
𝑅𝑇 0.0821 𝐿 ∗ 𝑎𝑡𝑚 ∗ 298𝐾
1𝑚𝑜𝑙 𝐶3𝐻8
𝑚𝑜𝑙 ∗ 𝐾
3𝑚𝑜𝑙 𝐶𝑂2
𝑃𝑉
1.00𝑎𝑡𝑚 ∗ 1.00𝐿
=
=
= 0.040873 𝑚𝑜𝑙 𝑂2 ∗
= 0.245 𝑚𝑜𝑙
𝐿
∗
𝑎𝑡𝑚
𝑅𝑇 0.0821
5 𝑚𝑜𝑙 𝐶3𝐻8
∗ 298𝐾
𝑚𝑜𝑙 ∗ 𝐾
Same mol created. Both are used up! 0.245mol CO2 created!
𝑉𝐶𝑂2
𝑛𝑅𝑇
=
=
𝑃
𝐿 ∗ 𝑎𝑡𝑚
∗ 298𝐾
𝑚𝑜𝑙 ∗ 𝐾
= 0.300𝐿 𝐶𝑂2
2.00𝑎𝑡𝑚
0.245 𝑚𝑜𝑙 ∗ 0.0821
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REACTIONS:
H2O
N2
CO2
• The reaction of solid dimethylhydrazine, (CH3)2N2H2, and liquefied dinitrogen
tetraoxide, has been investigated as a rocket fuel; the reaction produces gaseous
carbon dioxide, nitrogen and water vapor, which are ejected as exhaust gases. In a
controlled experiment solid dimethylhydrazine reacted with excess nitrogen tetroxide
with a final pressure of 2.50 atm, with a temperature of 400.0K. Find the pressure of
each gas. Type the total pressure into learning catalytics.
Hint: Write out and balance the equation.
Hint2: what do you think must be true of the ratio of the gases pressures? Remember n∝P
They must equal the mole ratios coefficients: 4 H2O: 2 CO2: 3N2
Hint3: What is the total pressure? What are the mol ratios of products?
H2O=(4/9)*(2.50 atm)=1.11atm
CO2=(2/9)*(2.50 atm)=0.556 atm
N2=(3/9)*(2.50atm)=0.834 atm
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• A 1.00 L sample of chlorine gas at 1.00 atm and 298 K reacts completely with 1.00 L of nitrogen gas and 2.00 L of oxygen gas at the
same temperature and pressure. There is a single gaseous product, which fills a 2.00L flask at 1.00 atm and 298 K. Use this information to
determine the following characteristics of the product. A) its empirical formula, B) its molecular formula, C) the most favorable Lewis
structure based on formal charge arguments (the central atom is an N atom); d) the molecular shape.
Hint 1: (a)“reacts
completely” in this question
means no limiting reagent,
everything reacts all the way
to zero.
Hint 2: (a) Remember V ∝ n at
constant pressure. Also don’t
forget the reactants are
diatomic gases.
Hint 3: (b) You know P, V, T of
the final reaction flask, what is
n?, what mass of reactants
were put in? Where did all
those reactants go. What is
the units of molecular mass?
A)
B)
Ratio of gas moles=ratio of gas volume: 1Cl2:1N2:2O2
Plan: find total grams, find total moles. Do grams/moles.
To find total grams: do the following for each reactant:
Then
Turn into g using molar mass
To find final total moles: use ideal gas law for final reaction flask.
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• A 1.00 L sample of chlorine gas at 1.00 atm and 298 K reacts completely with 1.00 L of nitrogen gas and 2.00 L of oxygen gas at the
same temperature and pressure. There is a single gaseous product, which fills a 2.00L flask at 1.00 atm and 298 K. Use this information to
determine the following characteristics of the product. A) its emperical formula, B) its molecular formula, C) the most favorable Lewis structure
based on formal charge arguments (the central atom is an N atom); d) the molecular shape.
Find total g
Put in by
doing each
reactant
separately:
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1.15g+2.90g+2.62g=6.67g total
• A 1.00 L sample of chlorine gas at 1.00 atm and 298 K reacts completely with 1.00 L of nitrogen gas and
2.00 L of oxygen gas at the same temperature and pressure. There is a single gaseous product, which fills a
2.00L flask at 1.00 atm and 298 K. Use this information to determine the following characteristics of the
(I’m not sure we’ll make it to this one, if we
product. A) its emperical formula, B) its molecular formula, C) the most favorable Lewis structure based on
don’t the answer will be posted online)
formal charge arguments (the central atom is an N atom); d) the molecular shape.
MORE REACTIONS:
Ratio of gas moles=ratio of gas volume: 1Cl2:1N2:2O2
Total Mass
(from other slide)
=6.67g total
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Total Moles reactant:
Compare to mass of NClO2=14+35+32= 81g/mol=
same, so NClO2 is also the molecular formula
• A 1.00 L sample of chlorine gas at 1.00 atm and 298 K reacts completely with 1.00 L of nitrogen gas and 2.00 L of oxygen gas at the same temperature and
pressure. There is a single gaseous product, which fills a 2.00L flask at 1.00 atm and 298 K. Use this information to determine the following characteristics of
the product. A) its emperical formula, B) its molecular formula, C) the most favorable Lewis structure based on formal charge arguments (the central atom is an
N atom); d) the molecular shape.
Steric number 3:
trigonal planar.
Why can’t we add another double bond to minimize formal charge further?
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