#11 - 25

**Problem #11:** How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mm Hg pressure?

**Solution:**

Rearrange the Ideal Gas Law to this:

n = PV / RT

Substitute values into the equation:

n = [(750.0 mmHg / 760.0 mmHg atm¯^{1}) (0.890 L)] / [(0.08206 L atm mol¯^{1}K¯^{1}) (294.0 K)]

Please note the division of 750 by 760. This is done in order to convert the pressure from mmHg to atm, because the value for R contains atm as the pressure unit. If we used mmHg, the pressure units would not cancel and we need to have them cancel because we require mol (and only mol) to be in the answer.

**Problem #12:** 1.09 g of H_{2} is contained in a 2.00 L container at 20.0 °C. What is the pressure in this container in mmHg?

**Solution:**

Rearrange the Ideal Gas Law to this:

P = nRT / V

Substitute values into the equation:

P = [(1.09 g / 2.02 g mol¯^{1}) (0.08206 L atm mol¯^{1}K¯^{1}) (293.0 K)] / 2.00 L

Please note the division of 1.09 by 2.02. This is done in order to convert grams to moles, because the value for R contains mol as the unit for amount of substance. If we used g, the mol unit in R would not cancel and we need to have it cancel because we require atm (and only atm) to be in the answer.

Multiply the answer (which is in atm) by 760.0 mmHg atm¯^{1} to get mmHg

**Problem #13:** Calculate the volume 3.00 moles of a gas will occupy at 24.0 °C and 762.4 mm Hg.

**Solution:**

Rearrange the Ideal Gas Law to this:

V = nRT / P

Substitute values into the equation:

V = [(3.00 mol) (0.08206 L atm mol¯^{1}K¯^{1}) (297.0 K)] / (762.4 mmHg / 760.0 mmHg atm¯^{1})

Note the conversion from mmHg to atm in the denominator.

**Problem #14:** How many moles of gas would be present in a gas trapped within a 100.0 mL vessel at 25.0 °C at a pressure of 2.50 atmospheres?

**Solution:**

Rearrange the Ideal Gas Law to this:

n = PV / RT

Substitute values into the equation:

n = [(2.50 atm) (0.1000 L)] / [(0.08206 L atm mol¯^{1}K¯^{1}) (298.0 K)]

**Problem #15:** How many moles of a gas would be present in a gas trapped within a 37.0 liter vessel at 80.00 °C at a pressure of 2.50 atm?

**Solution:**

Rearrange the Ideal Gas Law to this:

n = PV / RT

Substitute values into the equation:

n = [(2.50 atm) (37.0 L)] / [(0.08206 L atm mol¯^{1}K¯^{1}) (353.0 K)]

**Problem #16:** What volume will 1.27 moles of helium gas occupy at STP?

**Solution:**

1) Rearrange the Ideal Gas Law to this:

V = nRT / P

2) Substitute values into the equation:

V = [(1.27 mol) (0.08206 L atm mol¯^{1}K¯^{1}) (273.0 K)] / 1.00 atmor

(22.4 L / 1.00 mol) = (x / 1.27 mol) <--- only works at STP

3) Would it make any difference in the answer if the gas were oxygen? Krypton? Carbon dioxide? Methane?

No, no, no, no. The exact identity of the gas makes no difference to the number of moles present. By the way, note that, since the temperature and pressure would be the same, the same volume will contain the same number of molecules of gas, i.e. moles of gas. This is Avogadro's Hypothesis.

**Problem #17:** At what pressure would 0.150 mole of nitrogen gas at 23.0 °C occupy 8.90 L?

**Solution:**

P = nRT / VP = [(0.150 mol) (0.08206 L atm mol¯

^{1}K¯^{1}) (296.0 K)] / 8.90 L

**Problem #18:** What volume would 32.0 g of NO_{2} gas occupy at 3.12 atm and 18.0 °C?

**Solution:**

V = nRT / PV = [(32.0 g / 46.0 g mol¯

^{1}) (0.08206 L atm mol¯^{1}K¯^{1}) (291.0 K)] / 3.12 atm

Note the conversion from grams to moles (the 32.0/46.0).

**Problem #19:** How many moles of gas are contained in a 50.0 L cylinder at a pressure of 100.0 atm and a temperature of 35.0 °C? If the gas weighs 79.14 g, what is its molecular weight?

**Solution:**

n = PV / RTn = [(100.0 atm) (5.00 L)] / [(0.08206 L atm mol¯

^{1}K¯^{1}) (308.0 K)]n = 19.7828 mol (I kept some guard digits.)

molec. wt. = 79.14 g / 19.7828 mol = 4.00 g/mol (to 3 significant figures)

**Problem #20:** An amount of an ideal gas at 290.9 K has a volume of 17.05 L at a pressure of 1.40 atm. What is the pressure of this gas sample when the volume is halved and the absolute temperature is multipled by four?

**Solution:**

1) Calculate the moles of gas (I left off the units):

n = [(1.40) (17.05)] / [(0.08206) (290.9)] = 1.00 mol2) Use PV = nRT again, but divide the volume by two and multiply the temperature by 4. Use the mole value from just above and solve for P:

P = [(1.00) (0.08206) (290.9 x 4)] / (17.05/2) = 11.2 atm

There is another, more conceptual way, to explain the answer. First, we will think of the relationship between pressure and volume (they are inversely proportional). Then, we will think of the relationship between absolute temperatre and pressure (they are directly related).

If volume decreases by a factor of two, what happens to the pressure? The pressure INCREASES by a factor of two. So, your pressure would be doubled. P = 1.40 times 2 = 2.80 atm.

If the absolute temperature increases, pressure increases by the same amount. Therefore, since temperature is multiplied by 4, then pressure should be multiplied by 4. P = 2.80 times 4 = 11.2 atm

**Problem #21:** A balloon has a mass of 0.5 g when completely deflated. When it is filled with an unknown gas, the mass increases to 1.7 g. You notice on the canister of the unknown gas that it occupies a volume of 0.4478 L at a temperature of 50 °C. You note the temperature in the room is 25 °C. Identify the gas.

**Solution:**

1) Use Charles' Law to get volume at 25 °C:

0.4478 / 323 = x / 298

2) Assume balloon is at 1.00 atm. Use PV = nRT to get moles of gas:

(1.00) (0.41314) = (x) (0.08206) (298)

3) Divide grams of gas (1.7 g − 0.5 g = 1.2 g) by moles to get molec weight:

1.2 g / 0.01689466 mol = 71.0 g/molThe gas is chlorine, Cl

_{2}

By the way, the use of Charles' law in step one assumes that the entire contents of the gas cannister has been discharged. If we were to assume the cannister has only been partially discharged, then we cannot solve this problem.

**Problem #22:** A 10.20 g sample of a gas has a volume of 5.25 L at 23.0 °C and 751 mmHg. If 2.30 g of the same gas is added to this constant 5.25 L volume and the temperature raised to 67.0 degrees Celsius, what is the new gas pressure?

**Solution:**

1) Use PV = nRT to determine moles of 10.20 g sample:

(751 mmHg/760 mmHg atm^{-1}) (5.25 L) = (n) (0.08206 L atm mol^{-1}K^{-1}) (296 K)n = 0.21358 mol

2) Determine molecular weight of gas:

10.20 g / 0.21358 mol = 47.757 g/mol

3) Determine moles of 2.30 g of gas:

2.30 g / 47.757 g/mol = 0.04816 mol

4) Determine new pressure with new amount of moles and at new temperature:

(P) (5.25 L) = (0.26174 mol) (0.08206 L atm mol^{-1}K^{-1}) (340 K)P = 1.39 atm

Note: the 0.26174 came from 0.21358 plus 0.04816.

**Problem #23:** A gas consisting of only carbon and hydrogen has an empirical formula of CH_{2}. The gas has a density of 1.65 g/L at 27.0 °C and 734.0 torr. Determine the molar mass and molecular formula of the gas.

**Solution:**

1) Use PV = nRT to determine moles of the gas in 1.00 L:

(734.0 torr/760.0 torr atm^{-1}) (1.00 L) = (n) (0.08206 L atm mol^{-1}K^{-1}) (300. K)n = 0.039231 mol

2) Determine molecular weight of gas:

1.65 g / 0.039231 mol = 42.06 g/mol

3) Determine the molecular formula:

The "empirical formula weight" of CH_{2}= 14.027"Empirical formula weight" units of CH

_{2}present:42.06 / 14.027 = 2.998 = 3The molecular formula is C

_{3}H_{6}

Please be aware that "empirical formula weight" is not a standard term in chemistry.

**Problem #24:** 13.9 grams of an unknown gas is placed in a 5.00 L container.It has an initial pressure at 58.6 kPa and initial temperature at 60.0 °C. What is the name of this gas?

**Solution:**

1) Use PV = nRT:

(58.6 kPa) (5.00 L) = (n) (8.31447 L kPa mol^{-1}K^{-1}) (333 K)n = 0.10582 mol

I found the value for R here.

2) Determine molecular weight:

13.9 g / 0.10582 mol = 131.4 g/molExamining a periodic table leads us to identifying the gas as xenon.

**Problem #25:** A 19.5 L flask at 15 °C contains a mixture of three gases: N_{2} (2.50 mol), He (0.38 mol), and Ne (1.34 mol). Calculate the partial pressure of neon gas in the mixture.

**Solution:**

1) Determine total moles of gas:

2.50 + 0.38 + 1.34 = 4.22 moles

2) Use PV = nRT:

(x) (19.5 atm) = (4.22 mol) (0.08206) (288 K)x = 5.115 atm

Determine the partial pressure for neon:

5.115 x (1.34/4.22) = 1.62 atmNote: (1.34/4.22) determines the mole fraction of neon.

**Problem #26:** A 1.00 L flask is filled with 1.25 g of argon at 25.0 °C. Ethane vapor is then added to the same flask until the total pressure is 1.050 atm. What is the mass of ethane that was added to the flask?

**Solution:**

1) Determine the partial pressure of the Ar:

PV = nRT(P) (1.00 L) = (1.25 g / 39.948 g/mol) (0.08206 L atm / mol K) (298 K)

P = 0.76518 atm

2) Determine the ethane partial pressure:

1.050 atm − 0.76518 atm = 0.28482 atm

3) Determine moles of ethane added:

PV = nRT(0.28482 atm) (1.00 L) = (n) (0.08206 L atm / mol K) (298 K)

n = 0.011647 mol

4) Determine mass of ethane:

(0.011647 mol) (30.0694 g/mol) = 0.350 g (to three sig figs)

**Problem #27:** An experiment was performed to determine the value for R in the Ideal Gas Law. An excess of HCl(aq) was reacted with a piece of magnesium ribbon and the hydrogen gas released in the reaction was collected over water.

The following data were collected (100. cm of ribbon weighs 1.89 g):

P _{room}753.0 mmHg T _{room}22.7 °C V _{gas}30.00 mL Mg ribbon 1.55 cm

**Solution:**

**Comment:** The most commonly used unit on R is liter-atm / mol-K. That is the unit we will strive for in the solution.

1) The pressure of the hydrogen gas collected over water also includes a small amount of pressure from the water vapor. We will use Dalton's Law to calculate the pressure of just the hydrogen gas (referred to as the dry gas):

P_{tot}= P_{H2}+ P_{H2O}753.0 mmHg = P

_{H2}+ 20.7 mmHgP

_{H2}= 732.3 mmHgI used this table to calculate the vapor pressure. I interpolated between the 20.565 value and the 20.815 value to get 20.69 mmHg. From there, I rounded to the 20.7 value used in the calculation.

This last step will convert the mmHg to atm:

732.3 mmHg / 760.0 mmHg/atm = 0.9635526 atm

2) The temperature of the experimental apparatus and chemicals is assumed to be the same as the room temperature. We need that value in Kelvins:

K = °C + 273.15K = 22.7 + 273.15 = 295.85 K

3) The volume of the gas must be expressed in liters:

30.00 mL = 0.03000 L

4) The amount of magnesium will be expressed in moles. A ratio and proportion will be used first to determine the mass of the Mg ribbon:

100 cm 1.89 g ––––––– = ––––––– 1.55 cm x x = 0.029295 g

0.029295 g / 24.305 g/mol = 0.00120531 mol

5) We are now ready to calculate R:

PV = nRT

PV (0.9635526 atm) (0.03000 L) R = ––– = –––––––––––––––––––––––– = 0.08106 L-atm / mol-K nT (0.00120531 mol) (295.85 K)

6) In error by about 1.2% when using 0.08206 for the true value? Pretty good!

**Problem #28:** An ideal gas mixture containing molecular nitrogen and molecular hydrogen weighs 4.50 g and occupies a volume of 8.46 L at 300. K and 1.00 atm. Calculate the mass percent of these two gases in the mixture.

**Solution:**

1) Determine the moles of gas present:

PV = nRTn = [(1.00 atm) (8.46 L)] / [(0.08206 L atm / mol K) (300 K)]

n = 0.343651 mol

2) Assign x to be the mass of hydrogen in the mixture. Therefore, this is the mass of nitrogen in the mixture:

4.50 − x

3) The moles of each gas is as follows:

moles H_{2}---> x / 2.016

moles N_{2}---> (4.50 − x) / 28.014

4) Resulting in:

x 4.50 − x ––––––– + ––––––– = 0.343651 2.016 28.014

5) Clear the denominators:

(x) (2.016) (28.014) (4.50 − x) (2.016) (28.014) –––––––––––––––– + ––––––––––––––––––––– = (0.343651) (2.016) (28.014) 2.016 28.014 (28.014) (x) + (2.016) (4.50 − x) = 19.40811

6) Solve for x:

28.014x + 9.072 − 2.016x = 19.4081125.998x = 10.33611

x = 0.39757

7) Calculate the mass percents:

hydrogen ---> (0.39757 / 4.50) * 100 = 8.83%

nitrogen ---> 100 − 8.83 = 91.17%

**Bonus Problem #1:** The vapor pressure of water at 25 °C is 23.76 torr. If 1.50 g of water is enclosed in a 2.0 L container, will any liquid be present? If so, what mass of liquid?

**Solution:**

1) Use the ideal gas law to find out how many moles of gas would have to be vaporized to obtain a pressure of 23.76 torr.

PV = nRTP = gas pressure in atm = 23.76 torr x (1 atm / 760 torr) = 0.0313 atm

V = gas volume in L = 2.0

n = moles of gas = ?

R = gas constant = 0.08206 L atm / K mole

T = Kelvin temperature = 25 °C + 273 = 298 Kn = PV / RT = (0.0313)(2.0) / (0.08206)(298) = 0.00255992 moles of H

_{2}O gas

2) Determine mass of the water vapor:

0.00255992 moles H_{2}O gas x (18.015 g H_{2}O / 1 mole H_{2}O) = 0.046117 g H_{2}O gas

3) Amount of H_{2}O liquid in the container:

total g H_{2}O − g H_{2}O gas ---> 1.50 − 0.046117 = 1.45 g H_{2}O liquid (to three sig figs)

**Bonus Problem #2:** Container A holds N_{2} gas with a mass of 56.2 g and is 4.4 times the volume of container B which holds argon (Ar) gas at the exact same temperature and pressure. What is the mass of the Ar (in g) within container B?

**Solution:**

1) For this problem, there are two equations of interest:

PV = nRT

n = m/M, where M is the molar mass of the gas and m is the mass of the gas

2) Substituting one into the other, we have this:

PV = (m/M)RTrearrange it to this:

VM/m = RT/P

3) Some factors are constant, some are variable:

R is always a constant.

The problem specifies P and T are also constant.This means RT/P is constant.

4) Which means:

VM/m = constant

5) Since there is a VM/m for nitrogen and a VM/m for argon, we have this:

V_{1}M_{1}/m_{1}= V_{2}M_{2}/m_{2}cross multiply:

m

_{2}V_{1}M_{1}= m_{1}V_{2}M_{2}divide by V

_{1}M_{1}:m

_{2}= m_{1}(V_{2}/V_{1}) (M_{2}/M_{1})

6) Assign values and solve:

Container A (nitrogen): V_{1}= 4.4V_{2}, M_{1}= 28.0 g/mol

Container B (argon): V_{2}= V_{2}, M_{2}= 40.0 g/molMass of Ar in container B = (56.2 g) (V

_{2}/4.4V_{2}) (40.0/28.0) = 18.2 g (to 3 sig. fig.)

7) Comment: I could have assigned an arbitrary volume of 1 to V_{2}, making the value for V_{1} be 4.4. I could have done this since I know the volume of A (which is V_{1}) is 4.4 times the volume of B (which is V_{2}).