Assorted Gas Law Problems
Problems 1 - 10

Problems 11-25

Problems 26-50

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Problem #1:

A sample of CH4 is confined in a water manometer. The temperature of the system is 30.0 °C and the atmospheric pressure is 98.70 kPa. What is the pressure of the methane gas, if the height of the water in the manometer is 30.0 mm higher on the confined gas side of the manometer than on the open to the atmosphere side. (Density of Hg is 13.534 g/mL).  

Solution:

1) Convert 30.0 mm of H2O to equivalent mm of mercury:

(30.0 mm) (1.00 g/mL) = (x) (13.534 g/mL)

x = 2.21664 mm (I will carry some guard digits.)

2) Convert mmHg to kPa:
2.21664 mmHg x (101.325 kPa/760.0 mmHg) = 0.29553 kPa
3) Determine pressure of enclosed wet CH4:
At point A in the above graphic, we know this:

Patmo. press. = Pwet CH4 + Pthe 30.0 mm water column

98.70 kPa = x + 0.29553 kPa

x = 98.4045 kPa

5) Determine pressure of dry CH4:

From Dalton's Law, we know this:

Pwet CH4 = Pdry CH4 + Pwater vapor

(Water's vapor pressure at 30.0 °C is 31.8 mmHg. Convert it to kPa.)

98.4045 kPa = x + 4.23965 kPa

x = 94.1648 kPa

Based on provided data, use three significant figures; so 94.2 kPa.


Problem #2: 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) Let x = moles N2 and y = moles Ne. Therefore, from the problem:

28x + 20y = 10

2) However x = y, therefore:

48x = 10

x = 0.208 moles N2
y = 0.208 moles Ne

total moles in mixture = 0.416
"molar mass" of mixture = 10.0 g / 0.416 mol = 24.0 g/mol

3) Manipulate PV = nRT as follows:

a) replace n with g ÷ g mol¯1

b) PV = gRT / MM (MM is the molar mass)

c) PV (MM) = gRT

d) P (MM) / RT = g/V

(g/V is the unit for gas density.)

4) Insert values and solve:

(15.0 x 24.0) / (0.08206 x 500) = g/V

density = 8.8 g/L

Comment: using 28.014 and 20.18 rather than 28 and 20 (and carrying some guard digits) should refine the value a bit better. The book answer to this problem is 8.81 g/L.


Problem #3:

Three 1.00 L flasks at 25.0 °C and 1013 hPa pressure contain: CH4 (flask A), CO2 (flask B) and NH3 (flask C). Which flask (or none) contains 0.041 mol of gas?

Solution:

Three important points:

(a) All three flasks are at equal pressures (by the way, 1013 hPa is not usually seen. It is the same as 101.3 kPa, which is 1 atmosphere).

(b) All three flasks are at the same temperature.

(c) All three flasks have the same volume.

The above satisfies Avogaro's Hypothesis: equal volumes of gases under the same conditions of pressure and temperature, contain equal number of molecules.

Therefore, all three flasks either all contain 0.041 mol or none does. We will check flask A, using PV = nRT:

(1.00 atm) (1.00 L) = (n) (0.08206 L atm mol¯11) (298 K)

n = 0.0409 mol

All three flasks contain 0.041 mol of the different gases.


Problem #4: What is height (in mm) of a column of ethanol if the pressure at the base of the column is 1.50 atm? (The density of Hg is 13.534 g/cm3 and ethanol is 0.789 g/cm3.)

Solution:

1) convert atm to mmHg:

(1.50 atm) (760 mmHg/atm) = 1140 mmHg

2) convert mmHg to mm of ethanol:

(x) (0.789 g/cm3) = (1140 mmHg) (13.534 g/cm3)

x = 19,555 mmC2H5OH


Problem #5: What is height (in mm) of a column of methane if the pressure at the base of the column is 1.50 atm? (The density of Hg is 13.534 g/cm3 and methane is 0.717 kg/m3.)

Solution:

1) convert atm to mmHg:

(1.50 atm) (760 mmHg/atm) = 1140 mmHg

2) convert kg/m3 to g/cm3

kg/m3 x (1000 g/kg) = 1000 g/m3

1000 g/m3 x (m3/106 cm3) = g / 1000 cm3

In other words, to convert from kg/m3 to g/cm3, you divide by 1000

0.717 kg/m3 = 0.000717 g/cm3

3) convert mmHg to mmCH4:

(x) (0.000717 g/cm3) = (1140 mmHg) (13.534 g/cm3)

x = 2.15 x 107 mmCH4


Problem #6: 1.0 L of liquid nitrogen is kept in a closet measuring 1.0 m by 1.0 m by 2.0 m. Assuming that the container is completely full, that the temperature is 25.0 °C, and that the atmospheric pressure is 1.0 atm, calculate the percent (by volume) of air that would be displaced if all the liquid nitrogen evaporated. (Liquid nitrogen has a density of 0.807 g/mL.)

Solution:

1) calculate grams, then moles of N2:

0.807 g mL¯1 x 1000 mL = 807 g

807 g / 28.014 g mol¯1 = 28.8070 mol

2) Calculate volume of N2 at stated pressure and temperature:

V = nRT / P

V = [(28.8070) (0.08206) (298)] / 1.00

V = 704.44 L

3) Calculate volume of closet in liters:

1.0 m x 1.0 m x 2.0 m = 2.0 m3

2.0 m3 = 2000 L

Note: see here for short video on how to convert between m3 and L.

4) Assume N2 displaces 704.44 L of air:

704.44 / 2000 = 0.3522

35.2% of the air gets displaced.


Problem #7: A gaseous mixture of O2 and Kr has a density of 1.114 g/L at 385 torr and 400. K. What is the mole percent O2 in the mixture?

Solution:

1) Determine total moles of gas present:

PV = nRT

(385 / 760.) (1.00 L) = (n) (0.08206) (400.)

n = 0.0154332 mol

2) Two simultaneous equations in two unknowns are required. Let 'x' be the mass of oxygen and 'y' be the mass of Kr.

x + y = 1.114 g

(x / 32.00) + (y / 83.80) = 0.0154332 mol

3) Use 'x = 1.114 − y' and substitute into the second equation:

[(1.114 − y) / 32.00] + (y / 83.80) = 0.0154332

4) Multiply through by '(32.00) (83.80)' and continue to a solution:

[(32.00) (83.80)] [(1.114 − y) / 32.00] + [(32.00) (83.80)] (y / 83.80) = (0.0154332) (32.00) (83.80)

(83.80) (1.114 − y) + 32y = 41.38567

93.3532 − 83.8x + 32y = 41.38567

51.8y = 51.96753

y = 1.003 g (mass of Kr)

x = 0.111 g (mass of O2)

5) Determine total moles:

Kr ---> 1.003 g / 83.80 g/mol = 0.011968 mol
O2 ---> 0.111 g / 32.00 g/mol = 0.00346875 mol

0.011968 + 0.00346875 = 0.01543675 mol

6) Determine oxygen's mole fraction, then mole percent:

0.00346875 mol / 0.01543675 mol = 0.2247

0.2247 * 100 = 22.47% (by mole)


Problem #8: A humidifier is used in a bedroom kept at 22.0 °C. The bedroom's volume is 4.0 x 104 L. Assume that the air is originally dry and no moisture leaves the room while the humidifier is operating.

a. If the humidifier has a capacity of 3.00 gallons of H2O, will there be enough to saturate the room with water vapor (vp of H2O at 22. °C = 19.83 mmHg)?

b. What is the final pressure of water vapor in the room when the humidifier has vaporized two-thirds of its water supply?

Solution to part a:

1) Determine the mass of water vapor in the room (when saturated with water vapor):

PV = nRT; ⇒ n = PV/RT

n = (0.0261 atm) (4.0 x 104 L) / (0.08206) (295 K)

n = 43.11373 mol (I kept all the digits on my calculator. I skipped the units on R.)

grams = (43.11373 mol) x (18.0152 g/mol) = 776.7 grams of water

2) Determine the grams of water in 3.00 gallons:

One US gallon equals 3.78541178 L; ⇒ 3.00 gal = 11.35623534 L (We'll use 11.356 L)

11.356 L = 11,356 mL.

The density of water is 1.00 g/cm3 and 1 cm3 = 1 mL.

11,356 mL of water weighs 11,356 g.

The humidifier has sufficient capacity to saturate the room to the vapor pressure of water at 22.0 °C.

Solution to part b:

1) Two-thirds of the water supply is 7570.8 g of water.

2) Determine the pressure in the room:

PV = nRT; ⇒ P = nRT/V

P = [(420.246 mol) (0.08206) (295 K)[ / (4.0 x 104 L)

P = 0.254 atm (or 193.3 mmHg)


Problem #9: The vapor pressure of solid iodine at 30.0 °C is 0.466 mmHg.

a. How many milligrams of iodine will sublime into an evacuated 1.00 L flask?

b. If 2.00 mg of I2 are used, what will the final pressure be?

b. If 10.00 mg of I2 are used, what will the final pressure be?

Solution to part a:

1) Determine moles of I2:

PV = nRT; ⇒ n = PV/RT

n = [(6.1316 x 10¯4 atm) (1.00 L)] / [(0.08206) (303 K)]

n = 2.46603 x 10¯5 mol

2) Determine milligrams of I2:

grams of I2 = 2.46603 x 10¯5 mol x 253.8 g mol¯1 = 6.25878 x 10¯3 g

To 3 significant figures, this is 6.26 mg.

Solution to part b:

1) Determine moles of I2:

0.00200 g / 253.8 g mol¯1 = 7.88 x 10¯6 mol

2) Determine pressure:

PV = nRT; ⇒ P = nRT/V

P = [(7.88 x 10¯6 mol) (0.08206) (303 K)] / 1.00 L

P = 1.96 x 10¯4 atm (or 0.149 mmHg)

Another solution path uses this ratio and proportion: 6.26 is to 0.466 as 2.00 is to x. Cross-multiply and divide for the correct answer.

Solution to part c:

The pressure would be 0.466 mmHg. Remember that vapor pressure is an equilibrium process. At 0.466 mmHg there would be established an equilibrium between I2(s) and I2(g). There would be 3.74 mg of solid I2 remaining in the flask.

Problem #10: 20.0 g each of helium and an unknown diatomic gas are combined in a 1500. mL container. If the temperature is 298 K, and the pressure inside is 86.11 atm, what is the unknown gas?

Solution:

1) Determine the pressure exerted by the helium:

PV = nRT ⇒ P = nRT / V

x = [ (5.00 mol) (0.08206) (298 K) ] / 1.500 L

x = 81.513 atm

Comment: the 5.00 comes from 20.0 g / 4.00 g/mol. The 4.00 g/mol is the atomic weight of He.

2) Determine the pressure exerted by the unknown gas:

86.11 - 81.513 = 4.597 atm

3) Determine the moles of the unknown gas:

PV = nRT ⇒ n = PV / RT

x = [ (4.597 atm) (1.500 L) ] / [ (0.08206) (298 K) ]

x = 0.282 mol

4) Determine molecular weight and identity of gas:

20.0 g / 0.282 mol = 70.9 g/mol

The gas is chlorine, Cl2.


Bonus Problem: Cyanogen, a highly toxic gas, is composed of 46.2% C and 53.8% N by mass. At 25.0 °C and 750. torr, 1.05 g of cyanogen occupies 0.500 L. What is the molecular formula of cyanogen?

Solution:

1) Empirical formula for cyanogen:

Assume 100 g of cyanogen present.

C ---> 46.2 g / 12.0 g/mol = 3.85 mol
N ---> 53.8 g / 14.0 g/mol = 3.84 mol

The 1:1 molar ratio means the empirical formula is CN.

2) Moles of cyanogen in sample:
PV = nRT

(750. torr / 760. torr/atm) (0.500 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.0201776 mol

3) Molar mass of cyanogen:

1.05 g / 0.0201776 mol = 52.0 g/mol

4) Determine molecular formula:

The weight of CN is 12.0 + 14.0 = 26.0

52.0 / 26.0 = 2

The molecular formula for cyanogen is C2N2, usually written as (CN)2.


Problems 11-25

Problems 26-50

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