Chemistry 222OregonStateUniversity
Worksheet 3 Notes
1.A student reads the weather report and notices the pressure is listed as 29.77 inches. Express this pressure in mm of Hg. Express this pressure in atm.
29.77 inches = 756.2 mm Hg
756.2 mm Hg = 0.9950 atm
2.A metal can was imploded in lecture. Which of the following simplified laws best explains this behavior? P1V1=P2V2; V1/T1= V2/T2; V1/n1=V2/n2. Assume the variables not shown in each equation are held constant.
This is certainly up for individual interpretation, however, we believe V1/T1= V2/T2 is a good choice of description. The can is sealed so the moles of gas is a constant. As the temperature of the gas in the can decreased; the volume of the can decreased.
It can be declared that as the temperature of the gas in the can decreased; the pressure inside the can decreased; and the volume of the can decreased. So P1V1=P2V2.
3.Suppose we wish to contain a 1.00 mole sample of a gas at 1.00 atm pressure and 25 C. What volume container would we need?
PV = nRTV = nRT/PV = (1.00 mol)(0.0821 L•atm/mol•K)(298 K)/(1.00 atm)
V = 24.4 L
4.A gas at a temperature of 98.0 C (371 K) occupies a volume of 125.0 mL. To what temperature must the gas be lowered to occupy a volume of 100.0 mL?
PV = nRT
PV/nT = R = constant
=
P is a constant
n is a constant
=
125.0 mL = 100.0 mL
371 KT2
T2 = 296.8 K or 23.8 ºC
5.What is the density (g/L) of SF6 under standard conditions, 1.00 atm and 273K?
Assume one liter of gas (so the mass can be determined for one liter).
PV = nRT
n = PV/RT = (1.00 atm)(1.00 L)/(0.0821 L•atm/mol•K)(273 K) = 0.0446 moles SF6
0.0446 moles SF6 = 6.51 grams SF6
6.51 grams is the mass of one liter, so:
The density of SF6 is 6.51 g/L.
6.What is the pressure exerted by 30.0 g of CO2 (0.6817 moles) in a 20.00 L container at 500 C (773 K)? Give your answer in atmospheres, and mm of Hg, and psi (1 atm = 14.7 psi).
PV = nRTP = nRT/VP = (0.6817 moles)(0.0821 L•atm/mol•K)(773 K)/(20.00 L)
P = 2.16 atm
P = 2.16 atm 1641.6 mm Hg or 1.64 x 103 mm Hg
P = 2.16 atm 31.8 psi
7.A student obtains a 3.00 liter balloon at 23.0 ºC. She cools the balloon to –23.0 ºC. Calculate the volume of the balloon at –23.0 ºC.
PV = nRT
PV/nT = R = constant
=
P is a constant
n is a constant
=
3.00 L = V2 .
296.15 K250.15 K
V2 = 2.53 L
8.If 395 g NI3 detonates (about 14 oz.), how many liters of I2 vapor will be produced at 27 °C and 760 mmHg? (Hint: Balance equation first!)
NI3 (s) N2 (g) + I2 (g)
2 NI3 (s) N2 (g) + 3 I2 (g)
395 g_____ L
123
_____ mol_____ mol
Step 1:
395 g NI3 = 1.00 moles NI3 (s)
Step 2:
1.00 moles NI3 (s) = 1.50 moles I2 (g)
Step 3:
Use PV = nRT to solve for the volume of I2 (g). We just calculated that 1.50 moles I2 (g) are produced:
V = nRT/P = (1.50 moles)(0.0821 L•atm/mol•K)(300 K)/(1.00 atm) = 36.9 L
At what temperature should the same reaction be run in order to generate 67.2 L iodine gas from 395 g NI3 at standard pressure?
PV = nRT
T = PV/nR = (1 atm)(67.2 L)/(1.50 mol)(0.0821 L•atm/mol•K) = 546 K
In the reaction above, what volume would the iodine gas product occupy at 600 K?
PV = nRT
V = nRT/P = (1.50 mol)(0.0821 L•atm/mol•K)(600 K)/(1 atm) = 73.9 L
If 35.0 L of nitrogen gas is produced, how many liters of iodine gas will be produced?
2 NI3 (s) N2 (g) + 3 I2 (g)
35.0 L N2 = 105 L I2
9.What is the pressure of 4.0 mol of a gas in a 25-L container at standard temperature?
PV = nRTP = nRT/V = (4.0 mol)(0.0821 L•atm/mol•K)(273.15 K)/(25 L)
P = 3.6 atm
10.What is the temperature of 45.0 g Xe (0.343 mol) in a 5.0-L vessel at 916 Torr (1.21 atm)?
PV = nRTT = PV/nR = (1.21 atm)(5.0 L)/(0.343 mol)(0.0821 L•atm/mol•K)
T = 215 K
11.A student places 8.86 g of a noble gas into an 8.00-L container at 23.0 °C (296 K) and measures the pressure to be 512 mm Hg (0.674 atm). Identify this element.
PV = nRTn = PV/RT = (0.674 atm)(8.00 L)/(0.0821 L•atm/mol•K)(296 K)
n = 0.2219 moles of the unknown gas
Molar Mass = g/mol = 8.86 g/0.2219 moles = 39.9 g/mol (the noble gas having a molar mass of 39.9 is argon).
12.A 1-gallon milk jug (3.79 L) containing air at 1 atm and 25 °C (298.15 K) can withstand pressures up to 1.09 atm before the lid comes off. How high can it be heated before this happens?
PV = nRT
PV/nT = R = constant
=
V is a constant
n is a constant
=
1.00 atm = 1.09 atm
298.15 KT2
T2 = 325 K
13.A He balloon of 4.0 L at 1.00 atm floats into the air to an altitude where the pressure is 0.75 atm. What is its volume?
PV = nRT
PV/nT = R = constant
=
T is a constant
n is a constant
P1V1 = P2V2
(1.00 atm)(4.0 L) = (0.75 atm)(V2)
V2 = 5.33 L
14.A 10.0-L flask at a temperature of 300.0 K contains one mol of O2 gas, 2 moles of CO2 gas, and 7 moles of N2 gas. Calculate the total and partial pressures.
PV = nRT
P = nRT/V
The partial pressure of oxygen (the pressure inside the flask due to the oxygen gas molecules:
P(oxygen) = n(oxygen) RT/V
P(oxygen) = (1.00 mol)(0.0821 L•atm/mol•K)(300 K)/(10.00 L) = 2.46 atm
The partial pressure of carbon dioxide (the pressure inside the flask due to the carbon dioxide gas molecules:
P(oxygen) = n(carbon dioxide) RT/V
P(oxygen) = (2.00 mol)(0.0821 L•atm/mol•K)(300 K)/(10.00 L) = 4.93 atm
The partial pressure of nitrogen (the pressure inside the flask due to the nitrogen gas molecules:
P(oxygen) = n(nitrogen) RT/V
P(oxygen) = (7.00 mol)(0.0821 L•atm/mol•K)(300 K)/(10.00 L) = 17.2 atm
Total Pressure = 2.46 atm + 4.93 atm + 17.2 atm = 24.6 atm
15.A student burns 40.0-g propane (C3H8) in oxygen to produce carbon dioxide and water at STP. How many liters of steam are produced?
C3H8 (g) +5 O2 (g)→3 CO2 (g)+4 H2O (g)
40.0 g_____ L
123
_____ mol_____ mol
Step 1:
40.0 g C3H8 = 0.907 moles C3H8 (g)
Step 2:
0.907 moles C3H8 (g) = 3.63 moles H2O (g)
Step 3:
Use PV = nRT to solve for the volume of water. We just calculated that 3.63 moles H2O (g) are produced:
V = nRT/P = (3.63 moles)(0.0821 L•atm/mol•K)(273.15 K)/(1.00 atm) = 81.4 L
16.Does a neon-filled balloon float? (Hint: Why do things float?) Yes. Like a helium-filled balloon, a neon-filled balloon floats. The air that surrounds us is composed of 78% N2 (g) and 21% O2 (g) [the remaining 1% is Ar, CO2, H2O, CH4...] The molar mass of air is about 29 g/mol—this is due to N2 (g) being 28 g/mol and O2 (g) being 32 g/mol. The molar mass of Ne (g) is less (only 20 g/mol). So, the neon-filled balloon is less dense than air—it floats.
17.A couple of CH 121 students hop into a Toyota and drive to Eugene to visit an unenlightened friend. Using a few opportune reaction conditions, determine the liters of CO2 produced. What is the length (in feet) of a cubic container that would hold the CO2?
Molar Mass of Octane = 114.2285 g/mol /1.00000 Gallons = 3.78541 Liters
1.0000000 Liter = 0.03531467 cubic foot
Density of octane = 0.703 g/ml
Mileage Est. (mpg city/highway): 32/41
a.I will assume the combustion of 1.00 gallon of octane and the existence of
CO2 (g) at 298 and 1 atm when at equilibrium:
1.00 gal C8H18 = 3.785 L C8H18
b.3.785 L C8H18 = 3785 mL C8H18
c.3785 mL C8H18 = 2660 g C8H18
1.2660 g C8H18 = 23.3 mol C8H18
2.23.3 mol C8H18 = 186 mol CO2
d.PV = nRT
V = = = 4550 L CO2 (g)
[Change L to ft3 and discuss a cube that holds the carbon dioxide]:
4550 L CO2 (g) = 161 ft3 CO2 (g)
Volume of a cube = side3
Side = 3√V = 3√161 ft3
Side = 5.44 ft (5’ 5”)
18.What is the root-mean-square speed of H2 (use a molar mass of 2.0 g/mol or 0.0020 kg/mol) at 296 K and 1.00 atm (around room temperature and pressure)?
μrms= = = 1921 m/s
What is the root-mean-square speed of He (use a molar mass of 4.0 g/mol or 0.0040 kg/mol) at 296 K and 1.00 atm (around room temperature and pressure)?
μrms= = = 1359 m/s
Which gas is traveling faster? Why? H2 is traveling faster because it is less massive.
The mass you used for He was twice as great as that used for H2. Is He traveling twice as slow as H2? Why? He is traveling slower, but not twice as slow due to the square root function.
19.Consider the following six gases: CO (g) CO2 (g) Xe (g) He (g) F2 (g)SF6 (g)
Of these, which gas molecule has the greatest velocity at room temperature? Explain.
The gas molecule that has the greatest velocity is the lightest molecule (He).
Of these, which gas molecule has the lowest velocity at room temperature? Explain.
The gas molecule that has the lowest velocity is the heaviest (SF6 at 146 g/mol).
20.What is the mass of a 3.000 liter balloon that contains SF6 (g) at 1.01 atm and 296.0 K?
We are asked to determine the mass of gas present. Our equation of state (PV = nRT)
will enlighten us to the amount (in moles):
PV = nRTn = PV/RTn = (1.01 atm)(3.000 L)/(0.0821 L·atm/mol·K)(296.0 K)
n = 0.125 moles SF6 (g)
mass SF6 (g) = (moles)(g/mol) = (0.125 moles)(146.06 g/mol) = 18.2 grams SF6 (g)
21.Consider a sealed balloon containing nitrogen gas. Which of the following is false? Why?
(A)When the temperature is decreased, the velocity of the gas molecules decreases.
(B)When the temperature is decreased, the volume of the balloon decreases.
(C)When the temperature is decreased, the moles of gas inside the balloon decrease.
(D)When the temperature is decreased, the pressure inside the balloon stays constant.
(E)A 22.4-L balloon, at 1.00 atm, and 273.15 K contains one mole of nitrogen gas.
(C) is not correct because when the temperature is decreased the balloon is not compromised—no gas inside the balloon escaped. The other selections are true. When the temperature is decreased the gas molecules slow down, the balloon shrinks, the pressure inside the balloon continues to be the same as the pressure outside the balloon, and one mole of gas occupies 22.4 liters at 1.00 atm and 273 K.
22.A student burns 30.00 grams of hydrogen gas (H2) in oxygen to produce steam at 1.05 atm and 296 K. How many liters of steam are produced?
2 H2 (g) + O2 (g) → 2 H2O (g)
Step 1: The mass of hydrogen is given, change this into moles of hydrogen:
30.00 g H2 (g) = 14.88 mol H2 (g)
Step 2: Using the balanced chemical equation, determine the moles of steam produced from the moles of propane that will be consumed:
14.88 mol H2 (g) = 14.88 mol H2O (g)
Step 3: Using our equation of state (PV = nRT), determine the volume of steam from the other data (moles, pressure, temperature):
PV = nRTV = nRT/P = (14.88 mol)(0.0821 L·atm/mol·K)(296 K)/(1.05 atm)
V = 344.4 L