Assorted Gas Law Problems
Problems 26 - 50

Problems 1-10

Problems 11-25

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Problem #26: A container filled with 2.780 mol of N2, 3.990 mol of O2, and 13.42 mol of CO2, has a pressure of 1.252 atm.

(a) What is the mole fraction of each gas?
(b) What is the partial pressure of each gas?
(c) How big of a container would you need to fit these gases, if you were at 32.0 °C?

Solution:

1) Add up the moles:

2.780 + 3.990 + 13.42 = 20.19 mol

2) Calculate each mole fraction:

nitrogen ---> 1 − (0.1976 + 0.6647) = 0.1377
oxygen ---> 3.990 / 20.19 = 0.1976
carbon dioxide ---> 13.42 / 20.19 = 0.6647

3) Calculate each pressure:

nitrogen ---> 1.252 − (0.2473 + 0.8322) = 0.1725 atm
oxygen ---> (0.1976) (1.252 atm) = 0.2473 atm
carbon dioxide ---> (0.6647) (1.252 atm) = 0.8322 atm

4) Calculate volume:

PV = nRT

(1.252 atm) (V) = (20.19 mol) (0.08206 L atm / mol K) (305 K)

V = 403.6 L


Problem #27: At what temperature in Celsius will a 1.00 g sample of neon exert a pressure of 500. torr in a 5.01 L container?

Solution:

1) Determine moles of neon present:

1.00 g / 20.18 g/mol = 0.049554 mol

2) Use PV = nRT to determine the temperature in Kelvin:

(500. torr / 760. torr/atm) (5.01 L) = (0.049554 mol) (0.08206 L atm / mol K) (T)

T = 810. K

I converted from torr to atm because I have R in "L atm/mol K" memorized. I didn't want to bother looking up (or calculating) R in "L torr/mol K"

3) Convert to Celsius:
810. − 273 = 537 °C

Problem #28: If the argon atom has a radius of 1.54 Å, what percent of an argon gas sample at STP is actually empty space?

Solution:

1) Let's assume 1.00 mole of Ar is present. At STP, that sample occupies this volume:

22.414 L

You might recognize that as molar volume.

2) I'm going to convert 1.54 Å to cm:

(1.54 Å) (10¯8 cm / Å) = 1.54 x 10¯8 cm

3) Using the formula for volume of a sphere, let us determine the volume of one Ar atom:

V = (4/3) (3.14159) (1.54 x 10¯8 cm)3

V = 1.5298 x 10¯23 cm3

4) Determine the volume (in liters) occupied by 1.00 mole of the Ar atoms:

(1.5298 x 10¯23 cm3 / atom) (6.022 x 1023 atom¯1 = 9.2124556 cm3

9.2124556 cm3 = 0.0092124556 L

5) Determine the fraction of space occupied by Ar atoms:

0.0092124556 / 22.414 = 0.000411 (not a percent, but a decimal amount)

6) Determine the percentage of unoccupied space:

1 - 0.000411 = 0.999589

99.9589% (I decided to leave it unrounded off.)


Problem #29: A 2.5 L container filled with H2 at 468 mmHg is connected to a 3.5 L container containing Cl2 at 264 mmHg. The two containers are at the same temperature. A value (of negligible volume) between the two containers is opened and the gases react irreversibly, following this reaction:

H2(g) + Cl2(g) ---> 2HCl(g)

Assume that the reaction goes to completion and the temperature at the end is the same as at the beginning. What is the total pressure inside the two containers at the end of the reaction?

Solution:

The key factor is the total number of molecules does not change due to the reaction. For every one hydrogen reacting with one chlorine, two HCl are produced. The total number of particles does not change. Another way to express this is that the total number of moles of gas does not change.

Since moles and temperature remain constant, this problem now reduces to a Boyle's Law problem, involving only pressure and volume. The hydrogen container expands from 2.5 L (to 6.0 L) and the chlorine container expands from 3.5 L (to 6.0 L). The total pressure inside the 6.0 L is found via a Boyle's Law calculation:

P1V1 + P2V2 = P3V3

(468 mmHg) (2.5 L) + (264 mmHg) (3.5 L) = (x) (6.0 L)

x = 349 mmHg

You could also do this with two Boyle's law calculations and then add the results:

(468 mmHg) (2.5 L) = (x) (6.0 L)
(264 mmHg) (3.5 L) = (y) (6.0 L)

You would then add x and y to get 349 mmHg.


Problem #30: A 1.00 L container contains 0.20 g of H2. A 2.00 L container contains 8.0 g of X. The two containers are at the same pressure and temperature. Determine the molecular weight of X.

Solution:

1) For both gases, P, R, and T are constant. Let us rearrange:

PV = nRT

V = n (RT/P)

Let RT/P be represented by K

V = nK

2) Solve V = nK for K, using H2:

1.00 L = (0.10 mol) (K)

K = 10. L/mol

3) Solve V = nK for n, using gas X:

2.00 L = (n) (10. L/mol)

n = 0.20 mol

4) Determine molecular weight of X:

molec wt. = mass / n

8.0 g / 0.20 mol = 40. g/mol


Problem #31: What is the effect of the following change on the volume of 1 mol of an ideal gas? The initial pressure is 722 torr, the final pressure is 0.950 atm, the initial temperature is 32 °F, and the final temperature is 273 K.

Solution:

It turns out that 722 torr and 0.95 atm are the same pressure. Also, 32 °F is the same temperature as zero Celsius and 273 K is the same temp in the Kelvin scale.

Since pressure and temperature do not change, there is no volume change.

Here is the question on Yahoo Answers if you want to examine the five answers given there.


Problem #32: A gas has a volume of 2.80 L at 1.17 atm and 0 °C. At what temperature does it have a volume of 7.50 L at 517 mmHg?

Solution:

1) The units must match. Convert atm to mmHg:

1.17 atm times 760. mmHg / atm = 889.2 mmHg

2) Use the combined gas law:

P1V1T2 = P2V2T1

(889.2 mmHg) (2.80 L) (T2) = (517 mmHg) (7.50 L) (273 K)

T2 = 425 K

This is 152 °C


Problem #33: To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.60 L bulb, then filled it with the gas at 1.80 atm and 22.0 °C and weighed it again. The difference in mass was 9.50 g. Identify the gas.

Solution:

V = 4.60 L, P = 1.80 atm, T = 295 K, m = 9.50 g

Find the molar mass

PV = nRT ---> ideal gas equation

PV = mRT/M ---> n = m/M where m = mass, M = molar mass

M = mRT / (PV)

M = [(9.50 g) (0.08206 L atm/mol K) (295 K)] / [(1.80 atm) (4.60 L)]

M = 27.77 g/mol .... round to two significant digits .... 28 g/mol

The gas is nitrogen, N2.


Problem #34: A mixture of nitrogen and neon gases contains equal moles of each gas and has a total mass of 10.0 g. What is the density of this gas mixture at 500. K and 15.0 atm? Assume ideal gas behavior.

Solution #1:

1) Start off by working out how many moles of each gas you have in your mixture. Since they are both equal (but have different molar masses) you may write:

10 = 28.014x + 20.18x

x = 10/48.194 = 0.2075 moles of each gas

and 0.4150 moles in total.

2) Use V = nRT/P:

V = [(0.4150 mol) (0.08206 L atm / mol K) (500. K)] / 15.0 atm

V = 1.13516 L

3) Since density = mass/volume:

d = 10.0 g / 1.13516 L = 8.81 g/L

Solution #2:

1) We will manipulate PV = nRT. Let m = mass of gas in problem and M = molar mass of that gas. Therefore n = m / M and we have this:

PV = nRT

PV = mRT/M

2) Rearrange to get m/V, which is the density of the gas:

PM = mRT/V

3) Let d = density and since m/V equals density, substitutde "d" for m/V:

PM = dRT

4) Rearrange to isolate the density:

d = PM/RT
5) Let M = 24.097, insert values into symbolic equation and solve:
d = [(15.0 atm) (24.097 g/mol)] / [(0.08206 L atm / mol K) (500. K)]

d = 8.81 g/L

24.097 comes from the fact that there are equal moles of nitrogen and neon. This means there are equal nmbers of nitrogen molecules and neon atoms, so 24.097 is the average of 28.014 and 20.18.


Problem #35: The density of a gaseous compound containing carbon and hydrogen is found to be 0.716 g/L at STP. What is the molar mass of the compound? Speculate as to the identity of the compound.

Solution:

1) Rearrange PV = nRT as follows:

PV = (m/M)RT (m = mass; M = molar mass)

PM = (m/V)RT (m/V is the density of the gas, symbolized by 'd')

d = PM / RT

Since we want the molar mass:

M = dRT / P

2) Insert values and solve:

x = [(0.716 g/mol) (0.08206 L atm / mol K) (273 K)] / 1.00 atm

x = 16.0 g/mol

Methane, CH4, has this molar mass.


Problem #36: Ethylene oxide is produced from the reaction of ethylene and oxygen at 270-290 °C and 8-20 atm. In order to prevent potentially dangerous pressure buildups, the container in which this reaction takes place has a safety valve set to release gas when the pressure reaches 25.0 atm. If a 15.0 m3 reaction vessel contains 7.80 x 103 moles of gas, at what temperature will the pressure reach 25.0 atm? (There are 1000 L per cubic meter.)

Solution:

1) Determine liters:

15.0 m3 times 1000 L / m3 = 1.50 x 104 L

2) Use PV = nRT:

(25.0 atm) (15000 L) = (7800 mol) (0.08206 L atm / mol K) (x)

x = 586 K

This is 313 °C.


Problem #37: One flask contains 2.00 L of H2 at a pressure of 409 torr and is connected to a 1.00 L flask of N2 at an unknown pressure. The two flasks are connected to each other by a tube and a valve (both of negligible volume). The total pressure in the flasks is 340. torr after the valve between the two flasks is opened. Determine the initial pressure of N2 in the 1.00 L flask.

Solution:

P1V1 + P2V2 = P3V3

(409) (2) + (x) (1) = (340) (3)

818 + x = 1020

x = 202 torr


Problem #38: Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.5380 g/L at 25.0 °C and 721.0 mmHg?

Solution:

1) Determine the total moles of gas in the mixture:

PV = nRT

(721.0 mmHg / 760.0 mmHg/atm) (1.00 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.038795 mol

2) Let x = mass He and y = mass O2. Use two simultaneous equations in two unknowns:

x + y = 0.5380 g

(x / 4.0026) + (y / 31.9988) = 0.038795 mol

3) Substitue for y using y = 0.538 − x

(x / 4.0026) + [(0.5380 − x) / 31.9988)] = 0.038795

4) Clear the fractions by multiplying each term by (4.0026) * (31.9988):

(4.0026) (31.9988) (x / 4.0026) + (4.0026) (31.9988) [(0.5380 − x) / 31.9988)] = (0.038795) (4.0026) (31.9988)

(31.9988) (x) + (4.0026) (0.5380 − x) = 4.968957

5) And continue:

31.9988x + 2.153399 − 4.0026x) = 4.968957

27.9962x = 2.815558

x = 0.1006 g of helium

Therefore 0.4374 g of oxygen

6) Convert to moles:

He ---> 0.1006 g / 4.0026 g/mol = 0.02513 mol
O2 ---> 0.4374 g / 31.9988 g/mol = 0.01367 mol

7) Calculate the mole fraction, then mole percent, of helium:

0.02513 / (0.02513 + 0.01367) = 0.6477

mole percent ---> 0.6477 * 100 = 64.77% (by mole)

Here is a brief tutorial on mole percent.


Problem #39: A rigid stainless steel chamber contains CH4 at 140. torr of pressure and excess oxygen, O2, at 170.0 °C. A spark is ignited inside the chamber, completely combusting the methane. What is the change in total pressure within the chamber following the reaction? Assume a constant temperature throughout the process.

Solution:

CH4(g) + 2O2(g) ---> CO2(g) + 2H2O(g)

1) Here's the short explanation:

for every three moles of reactants (one CH4 and two O2) used up, three moles of products (one CO2 and two H2O) come to be made. No change in total amount of gaseous substance means no change in the pressure at reaction's end.

2) Here's the longer explanation:

140. torr of CH4 pressure means at least 280 torr of O2 pressure. Since it is excess, let us call the O2 pressure 280 + x.

Let us consider the CH4 to CO2 molar ratio of 1 to 1. Since this happens at constant volume and constant temperature, the pressures are directly related to the moles used and produced. Thus, we conclude 140. torr of CO2 pressure is produced.

Since we are at 170 °C, the water produced is gaseous. Let us consider the CH4 to H2O molar ratio of 1 to 2. This informs us that 280 torr of water vapor pressure is created.

Since O2 is in excess, we know that 'x' amount of pressure remains after 280 torr of O2 is used up.

Conclusion: no pressure change in the chamber at reaction's end.


Problem #40: In order for a gas-filled balloon to rise in air, the density of the gas in the balloon must be less than that of air. Consider air at 25.0 °C and 1.00 atm to have a molar mass of 28.96 g/mol. Determine the minimum temperature to which the balloon filled with oxygen at 1.00 atm would have to be heated before it could begin to rise in air. (Ignore the mass of the balloon itself.)

Solution:

1) I'm going to manipulate PV = nRT to get an expression equal to the density of the gas -- the (m/V) below:

PV = nRT

n = mass / molar mass = m / Mr

PV = (m/Mr) (RT) = mRT/Mr

P = (1/V) (mRT/Mr) = (m/V) (RT/Mr)

m/V = (P) (Mr/RT) = PMr/RT

2) The balloon will begin to rise at a temperature just above this condition:

density of air = density of oxygen

(m/V) of air = (m/V) of oxygen

3) To do this, we substitute into the above equation as follows:

(Pair) (Mr, air)/(RTair) = (PO2) (Mr, O2)/(RTO2)

Mr, air / Tair = Mr, O2 / TO2

Note how both P and R drop out.

TO2 = Tair times [Mr, O2 / Mr, air ]

4) The numerical solution:

TO2 = 298.15 times (31.9994/28.96)

TO2 = 329.44 K = 56.29 °C

As soon as the oxygen-filled balloon exceeds 56.29 °C, it will begin to rise.


Problem #41: 6.87 grams of gas occupy 3.45 L at 25.0 °C at 88.8 kPa. How many moles of gas is it? What is the molecular mass of the gas?

Solution:

1) Use PV = nRT

(88.8 kPa) (3.45 L) = (n) (8.31447 L kPa / mol K) (298 K)

n = 0.123646 mol (I kept some guard digits.)

2) Determine molecular weight:

6.87 g / 0.123646 mol = 55.6 g/mol (to three sig figs)

Comment: I use the value for R obtained from this page. It's in the second table, about half-way down the second column. There is the occasional teacher that INSISTS on you using the value for R that is 0.08206 L atm / mol K. In that case, convert the kPa unit to atm in this fashion:

88.8 kPa / 101.325 kPa/atm = 0.876388 atm

Problem #42: 17.3 grams of gas occupies 1096 mL at 85.0 °C and 192 kPa. What is the molecular mass of the gas?

Solution:

1) Use PV = nRT

(192 kPa / 101.325 kPa/atm) (1.096 L) = (n) (0.08206 L atm / mol K) (358 K)

n = 0.0706937 mol (I kept some guard digits.)

2) Determine molecular weight:

17.3 g / 0.0706937 mol = 245 g/mol (to three sig figs)

Comment: in this problem I used 0.08206 L atm / mol K for the value of R and this factor:

(192 kPa / 101.325 kPa/atm)

converts the pressure from kPa to atm.


Problem #43: What is the molecular mass of a gas which has a density of 2.78 g/L at 75.0 °C and 940.0 mmHg?

Solution:

1) We will use PV = nRT to determine the moles of gas in 1.00 L:

(940.0 mmHg / 760.0 mmHg/atm) (1.00 L) = (n) (0.08206 L atm / mol K) (348 K)

n = 0.0433115 mol

I used 1.00 L because gas density is measured in grams per liter, that is, to say grams per 1.00 L.

2) Determine molecular mass:

2.78 g / 0.0433115 mol = 64.2 g/mol

3) A slightly different approach manipulates the ideal gas equation before numbers are inserted, as follows:

PV = nRT

PV = mRT/M ---> n = m/M where m = mass, M = molar mass

M = mRT / (PV)

M = [(2.78 g) (0.08206 L atm/mol K) (348 K)] / [(1.23684 atm) (1.00 L)]

M = 64.2 g/mol

Note that I changed to pressure to atm before doing the calculation, If you had wanted to keep the 940 mmHg, then you would have needed an R with units of 'L mmHg / mol K.'


Problem #44: What is the density of ethane gas, C2H6, at STP?

Solution:

1) Determine moles in 1.00 L of ethane at STP:

(1.00 atm) (1.00 L) = (n) (0.08206 L atm / mol K) (273 K)

n = 0.044638 mol

2) Determine density in g/L:

30.0694 g/mol times 0.044638 mol/L = 1.43 g/L (to three sig figs)

30.0694 g/mol is the molecular mass of ethane.


Problem #45: Consider two steel tanks filled with the same mass of two different gases at the same pressure and temperature. One container is filled with Argon gas and has a volume of 10 liters. The other container is filled with a different gas and has a volume of 20 liters. Which of the following could be the other gas? (a) N2, (b) He, (c) Ne, (d) Cl2, (e) CO2

Solution:

1) We will use Avogadro's Hypothesis to solve this problem:

equal volumes of gas at equal temperature and pressure contain equal number of molecules.

2) The equal temperature and equal pressure are specified. Now, consider the volumes. If they were equal volumes, we would say that there were equal number of gas molecules. However, one volume is double the other. That means that the 20 L volume contains twice the number of particles as the 10 L volume. (If it helps, think of the 20 L volume as being two 10 L volumes and then do the comparison.)

3) We know that the two volumes have the same mass and that one is filled with Ar (molar mass = 40 g/mol). In order for the 20 L (with twice the number of gas molecules) to have the same mass as the 10 L, the mass of each individual gas molecule (in the 20 L) would have to be half the value of the argon.

4) Neon at 20 g/mol is the correct answer.


Problem #46: When 20 cm3 of gaseous alkene burns in an excess of oxygen, 60 cm3 of carbon dioxide is formed. When both volumes are measured at RTP, what is the formula of the alkene?

Solution:

1) Since everything occurs at the same temperature and pressure (that's the RTP), Avogadro's Hypothesis comes into play:

Equal volumes at equal T and P contain equal numbers of molecules.

This means 2 moles of the alkene produces 6 moles of CO2

This means the following equation:

CnH2n + O2 ---> 3CO2 + H2O

n must equal 3, therefore:

C3H6

Problem #47: True or False: Three spheres having the same volume are filled with gases at the same temperature and pressure. The gas in sphere A is pure hydrogen, the gas in sphere B is pure carbon monoxide, but the gas in sphere C is a mixture of hydrogen and carbon monoxide. Based on the gas laws and molecular formulas, all three spheres contain the same number of molecules.

Solution:

1) State Avogadro's Hypothesis:

Equal volumes at equal temperature and pressure contain equal number of molecules

2) We are informed by the problem that the volumes are equal. In addition, we are also informed that the temperature and the pressure are equal across the three spheres.

3) Therefore, the three spheres have equal number of molecules and the statement is shown to be true.


Problem #48: Would three balloons, each containing the same number of molecules of a different gas at STP have the same mass or same volume?

Solution:

1) Three of the four factors of Avogadro's Hypothesis are present:

same temperature? Check.
same pressure? Check.
same number of molecules? Check.

2) That means the fourth factor is also present:

same volume

Problem #49: A gas sample is found to be 40.1% sulfur and 59.9% oxygen. In a lab experiement you obtain the following data:

P = 1.00 atm; V = 500.0 mL; T = 298K

and the sample had a mass of 1.635g. Determine the molecular formula for the gas.

Solution:

1) Determine empirical formula of the gas:

sulfur ---> 40.1 g / 32.06 g/mol = 1.251 mol
oxygen ---> 59.9 g / 16.00 g.mol = 3.744 mol

sulfur ---> 1.251 mol / 1.251 mol = 1
oxygen ---> 3.744 mol / 1.251 mol = 2.99 = 3

The empirical formula is SO3.

2) Rearrange PV = nRT to determine moles of gas present:

n = PV / RT

n = [(1.00 atm) (0.5000 L)] / (0.08206 L atm / mol K) (298 K)]

n = 0.020447 mol

3) Determine molecular weight of gas:

1.635 g / 0.020447 mol = 79.96 g/mol

4) Determine molecular formula:

The weight of the empirical formula (SO3) is 80.062 g

79.96 / 80.062 = 0.9987 = 1

The molecular formula is also SO3


Problem #50:

Nitrogen can react with steam to form ammonia and nitrogen monoxide gas. A 20.0 L sample of nitrogen at 173 °C and 772 mmHg is made to react with an excess of steam. The products are collected at room temperature (25.0 °C) into an evacuated flask with a volume of 15.0 L.

(a) Write a balanced equation for the reaction.
(b) What is the total pressure of the products in the collecting flask after the reaction is complete?
(c) What is the partial pressure of each of the products in the flask?

Solution:

Part (a):

5N2 + 6H2O ---> 4NH3 + 6NO

Part (b):

moles N2 using PV = nRT ---> (772/760) (20.0) = (n) (0.08206) (446) = 0.5551 mol

I used (772/760) to convert from mmHg to atm.

There is a 5 : 10 (reduced, it is a 1 : 2 ratio) molar ratio between N2 and the products.

1 is to 2 as 0.5551 mol is to x

x = 1.1102 mol <--- moles of total products

total pressure using PV = nRT ---> (y) (15.0 L) = (1.1102 mol) (0.08206 L atm / mol K) (298 K) = 1.8099 atm

to three sig figs, 1.81 atm

Part (c):

There is a 5 : 4 molar ratio between N2 and NH3

5 is to 4 as 0.5551 mol is to z

z = 0.44408 mol <--- moles of ammoina

1.81 atm times (0.44408 mol / 1.1102 mol) = 0.724 atm <--- partial pressure of ammonia

(0.44408 mol / 1.1102 mol) is the mole fraction of ammonia

1.1102 atm − 0.724 atm = 0.386 atm <--- partial pressure of NO (to three sig figs)


Problems 1-10

Problems 11-25

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