Problems 11 - 20
Go to the first ten AP Gas Law Problems
Problem #11: The mean molar mass of the atmosphere at the surface of Titan, Saturn's largest moon is 28.6 g/mol. Titan's surface temperature is 95 K and its pressure is 1.6 atm. Assuming ideal behavior, calculate the density of Titan's atmosphere under these conditions.
Solution:
1) Let us assume the presence of one mole of gas. Determine its volume under the conditions of Titan's atmosphere:
PV = nRT2) Calculate the density:(1.6 atm) V = (1.00 mol) (0.08206 L atm / mol K) (95 K)
V = 4.8723125 L (I kept some guard digits)
28.6 g / 4.8723125 L = 5.87 g/L
Problem #12: The mean molar mass of the atmosphere at the surface of the Earth is 29.0 g/mol. Earth's surface temperature is 298 K and its pressre is 1.00 atm. Assuming ideal behavior, calculate the density of Earth's atmosphere under these conditions.
Solution:
1) Let us assume the presence of one mole of gas. Determine its volume under the conditions of Earth's atmosphere:
PV = nRT2) Calculate the density:(1.00 atm) V = (1.00 mol) (0.08206 L atm / mol K) (298 K)
V = 24.45388 L (I kept some guard digits)
29.0 g / 24.45388 L = 1.186 g/L
Comment: Titan's atmosphere is five times more dense than Earth's atmosphere.
Problem #13: A gas mixture composed of helium and argon has a density of 0.704 g/L at a 737 mmHg and 298 K. What is the percent composition of the mixture by (a) mass and by (b) volume.
Solution to a:
1) Calculate total moles of gases present:
Comment: assume that the volume of the gas mixture is 1.00 LPV = nRT ⇒ n = PV / RT
n = (737 torr / 760 torr/atm) (1.00 L) / (0.08206 L atm/mole K) (298 K)
n = 0.039656 mol
2) Set up two simultaneous equations:
Comment: let x = mass He and y = mass ArEquation #1 ⇒ x + y = 0.704 g
Equation #2 ⇒ x/4.0026 g/mol + y/39.948 g/mol = 0.039656 mol
3) Substitute x = 0.704 - y into the second equation and solve for y:
Comment: I left off the units until the end.(0.704 - y)/4.0026 + y/39.948 = 0.039656
(39.948) (0.704 - y) + (4.0026) (y) = (0.039656) (39.948) (4.0026)
More algebra results in:
y = 0.606 g Ar
x = 0.098 g He
4) Calculate mass percent of each gas:
Ar ⇒ (0.606 / 0.704) x 100 = 86.61%
He ⇒ 14.06%
Solution to b:
1) Let us determine the volume 0.606 g of Ar occupies at the stated T and P:
(737/760) (x) = (0.606/40) (0.08206) (298)x = 0.382 L
2) Since combined volume was 1.00 L, the volume percents are:
Ar ⇒ 38.2%
He ⇒ 61.8%
Problem #14: A sample of gas (1.90 mol) is in a flask at 21.0 °C and 697.0 mm Hg. The flask is now opened and more gas is added to the flask. The new pressure is 795.0 mm Hg and the temperature is now 26.0 °C. How many moles of gas are now in the flask?
Solution #1:
1) Use PV = nRT with the first set of data to get the volume of the container:
(697.0/760.0) (x) = (1.90 mol) (0.08206 L atm/mol K) (294.0 K)x = 49.9819572 L
2) Use PV = nRT with the second set of data, using the volume just calculated. Solve for moles:
(795.0/760.0) (49.9819572 L) = (x) (0.08206 L atm/mol K) (299.0 K)x = 2.13 mol
Solution #2:
1) Set up two uses of PV = nRT:
a) (0.9171 atm) (V1) = (1.90 mol) (R) (294.0 K)b) (1.046 atm) (V1) = (1.90 + x mol) (R) (299.0 K)
V1 represents the volume of the flask, which does not change.
2) Since V1 = V1 and R = R, divide (a) by (b):
(0.9171 atm / 1.046 atm) = [(1.90 mol) (294.0 K)] / [(1.90 + x mol) (299.0 K)]3) Total moles of gas in the flask:(0.9171) (1.90 + x) (299.0) = (1.046 atm) (1.90) (294.0)
521.00451 + 274.2129x = 584.2956
x = 0.23 mol
This is the amount of moles of gas added, not the total moles.
1.90 + 0.23 = 2.13 mol
Problem #15: In an experiment 350.00 mL of hydrogen gas was collected over water at 25.0 °C and 720.00 mmHg. Then, one-third of the gas leaked out of the container. What would the new volume be?
Solution:
1) Remove vapor pressure of water:
720.00 mmHg - 23.76 mmHg = 696.24 mmHg
2) Calculate moles of gas:
PV = nRT(696.24/760.00) (0.3500) = (n) (0.08206) (298)
n = 0.013111901 mol
3) Allow one-third to escape:
0.013111901 mol x (2/3) = 0.008741267 mol
4) Assume gas is still over water. Calculate new volume:
(720.00/760.00) (V) = (0.008741267) (0.08206) (298)V = 0.22563 L = 225.63 mL
Notice that I did not reduce the vapor pressure value by one-third. All vapor pressures are independent of the actual volume above the liquid. They are dependent only on the temperature.
Problem #16: What volume of SO2 at 25.0 °C and 1.50 atm contains the same number of molecules as 2.00 L of chlorine gas measured at STP?
Solution:
1) Calculate moles of Cl2:
PV = nRT(1.00) (2.00) = (x) (0.08206) (273.0)
x = 0.089276229 mol (I kept some guard digits.)
Since moles is a direct measure of the number of molecules, we do not have to determine how many molecules this is.
2) Determine volume of SO2 that holds 0.089. . . moles:
(1.50) (x) = (0.089276229) (0.08206) (298.0)x = 1.46 L
Problem #17: A mixture of CO2 and Kr weighs 35.0 g and exerts a pressure of 0.708 atm in its container. Since Kr is expensive, you wish to recover it from the mixture. After the CO2 is completely removed by absorption with NaOH(s) the pressure in the container is 0.250 atm. How many grams of CO2 and how many grams of Kr were initially present?
Solution: 1) Calculate the mole fractions of CO2 and Kr:
CO2: 0.458 atm / 0.708 atm = 0.64692) Change these to "gram fractions:"
Kr: 0.3531
CO2: 0.6469 x 44.0 = 28.4636
Kr: 0.3531 x 83.8 = 29.59CO2: 28.4636 /58.05 = 0.49
Kr: 0.51
Please be aware that "gram fractions" is not a standard term.
3) Calculate grams in the mixture:
CO2: 35.0 x 0.49 = 17.15 g
Kr: 17.85 g
Comment: Based on this ratio (0.458/0.250) the CO2:Kr molar ratio is 1.83 to 1. Is the above gram ratio also a 1.83 to 1 molar ratio?
Yes. Convert 17.15 g and 17.85 g to their respective moles and divide moles of CO2 by moles of Kr.
Problem #18: Which of the following is constant for 1 mole of any ideal gas?
a) PVT
b) PV/T
c) PT/V
d) VT/P
Solution:
PV = nRTPV/T = nR
Since the right-hand side is constant, the answer is B.
Problem #19: Our atmosphere is a mixture of gases (roughly 79% N2, 20% O2 and 1%Ar).
(a) What is the partial pressure (in atm) of each gas in the atmosphere?
(b) A mixture of He and O2 gases is used by deep sea divers. If the pressure of the gas a diver inhales is 8.0 atm what percent of the mixture should be O2, if the partial pressure of O2 is to be the same as what the divers would ordinarily breathe at sea level?
Solution to (a):
1) Determine the mole fraction of each gas. Assume 100 g of atmosphere:
N2: 79 g / 28.0 g mol¯1 = 2.82 mol
O2: 20 g / 32.0 g mol¯1 = 0.625 mol
Ar: 1 g / 40 g mol¯1 = 0.025 mol
2) Determine mole fraction of oxygen:
0.625 mol / 3.47 mol = 0.180
3) Determine partial pressure of oxygen:
1.00 atm x 0.180 = 0.180 atm
Solution to (b):
1) Calculate mole fraction of He/O2 mixture:
0.180 / 8 = 0.0225 mol of O2
7.820 / 8 = 0.9775 mol of He
2) Convert each to grams:
O2: 0.0225 mol x 32.0 g mol¯1 = 0.72 g
He: 0.9775 mol x 4.00 g mol¯1 = 3.91 g
3) Calculate percent of O2 in the mixture:
0.72 g / 4.63 g = 0.1555 = 15.55%
Problem #20: 600.0 mL of a mixture of O2 and O3 weighs 1.00 g at NTP. Calculate the volume that the ozone in mixture would occupy at NTP if it were alone.
Before the solution, a comment. I will take the N in NTP to mean 'normal,' with a synonym being 'room,' as in RTP. These values are taken to be 1.00 atm and 25.0 °C.
Solution:
1) Use PV = nRT to calculate the number of moles of gas:
(1.00 atm) (0.6 L) = (x) (0.08206) (298 K)x = 0.024536 mol
2) Determine grams of oxygen and ozone in the mixture:
x/32 + (1-x)/48 = 0.024536multiply each term by 32 to get:
x + 2/3 - (2/3x) = 0.785152
1/3 x = 0.118485
x = 0.355 g of O2 (to three sig figs)
So, ozone = 0.645 g
3) Calculate moles, then volume of ozone at NTP:
0.645 g / 48.0 g/mol = 0.0134375 mol(1.00 atm) (x) = (0.0134375 mol) (0.08206) (298 K)
x = 0.329 L = 329 mL
Problem #21: A mixture of Ar and N2 gases has a density of 1.419 g/L at STP. What is the mole fraction of each gas?
Solution:
1) At STP, the following is true:
a) the volume of one mole of gas occupies 22.414 L
b) the mass of one mole of Ar is 39.948 g
c) the mass of one mole of N2 is 28.014 g
2) Determine the mass of 22.414 L of the gas mixture:
1.419 g/L times 22.414 L = 31.805466 g
3) The mass of the gas mixture is the weighted average of the molar masses (since the molar masses occupy 22.414 L):
31.805466 = (x) (39.948) + (1 - x) (28.014)31.805466 = 39.948x + 28.014 - 28.014x
3.791466 = 11.934x
x = 0.3177
x is the mole fraction for Ar; the mole fraction for nitrogen is 0.6823.
Problem #22: You are given an envelope containing a piece of magnesium and a piece of zinc with a total mass of 0.0833 grams. The volume of gas collected (at 25.0 °C and Patm = 755 torr) is 59.74 mL. Liquid column height is only 15 mm. Calculate the mass of each piece of metal in the envelope.
Comment: I'm going to assume the hydrogen gas produced was collected over mercury. Assuming the gas was collected over water makes it a bit more complicated.
Solution:
1) Get the pressure in the gas collection tube:
755 torr minus 15 torr = 740 torr
2) Determine moles of gas produced:
PV = nRT(740 torr / 760 torr/atm) (0.05974 L) = (n) (0.08206) (298 K)
n = 0.0023787 mol
3) Set up first equation (of two):
m + z = 0.0833where m = mass of Mg and z = mass of Zn
4) Set up second equation:
(m / 24.305) + (z / 65.38) = 0.0023787m / 24.305 = moles of magnesium
z / 65.38 = moles of zincThe sum of the moles of Mg and Zn equals the total moles of gas collected because of the 1:1 stoichiometry of each reaction:
Zn (s) + H2SO4 (aq) ---> H2 (g) + ZnSO4 (aq)
Mg (s) + H2SO4 (aq) ---> H2 (g) + MgSO4 (aq)
5) Solve the two equations:
Here is the solution via a Cramer's rule (the method of determinants) on-line calculator:m = 0.04273415 g
z = 0.04056585 gTo three sig figs:
m = 0.0427 g
z = 0.0406 gIf you use the on-line app, make sure to use these values for the second equation:
m / 24.305 = 0.0411438m
z / 65.38 = 0.015295zAs an additional exercise, you may wish to solve the equation system by hand.