## About 15 billion years ago a tremendous explosion started the expansion of the

universe. This explosion is known as the Big Bang. At the point of this event

all of the matter and energy of space was contained at one point. What exsisted

prior to this event is completely unknown and is a matter of pure speculation.

#### This occurrence was not a conventional explosion but rather an event filling all

of space with all of the particles of the embryonic universe rushing away from

each other. The Big Bang actually consisted of an explosion of space within

itself unlike an explosion of a bomb were fragments are thrown outward. The

galaxies were not all clumped together, but rather the Big Bang lay the foundations for the universe.

### The Universe – Big Bang Nucleosynthesis

#### Helium and Deuterium, Oxygen and Hydrogen and traces of lithium and beryllium were some of the Gas particles formed. The amounts of these gas particles can be calculated given the volume and pressure at certain given constants. See sample problems below, related to different planets.

## The notion that the expanding universe was extremely hot in the beginning provides a reasonable explanation for why helium and deuterium seem to have existed even before star formation. Both these species are created by nuclear fusion. Fusion of a proton and a neutron produces deuterium (also known as heavy hydrogen), while fusion of two deuterium nuclei produces helium. These reactions can occur only at very high temperatures, such as in the interiors of stars. In 1946, George Gamow, once a student of Friedmann, suggested that nuclear fusion must have taken place when the universe was so hot in the beginning. This process, called the “big bang nucleosynthesis”, would have created helium and deuterium (plus trace amounts of elements like lithium and beryllium) out of an initial sea of energetic protons and neutrons.

### In the early 1960s, spectroscopic studies of local stars showed that the abundance of helium was about 20-30% by mass, the rest being mostly hydrogen. Stars and hydrogen bombs are the only things we know of that make helium in the present universe.

## They both combine hydrogen nuclei (protons) into helium nuclei through nuclear fusion, releasing great amounts of energy. Astronomers calculate that the night sky should be much brighter if all the helium we now observe had come from stars burning (or bombs exploding). Some, if not most, of the helium must have existed before star formation.

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 = nRT

(1.6 atm) V = (1.00 mol) (0.08206 L atm / mol K) (95 K)

V = 4.8723125 L (I kept some guard digits)

2) Calculate the density:

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 = nRT

(1.00 atm) V = (1.00 mol) (0.08206 L atm / mol K) (298 K)

V = 24.45388 L (I kept some guard digits)

2) Calculate the density:

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 L

PV = 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 Ar

Equation #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)]

(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.

3) Total moles of gas in the flask:

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 SO_{2} at 25.0 °C and 1.50 atm contains the same number of molecules as 2.00 L of chlorine gas measured at (Standard Temperature and Pressure) STP?

Solution:

1) Calculate moles of C_{l2}:

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 SO_{2} that holds 0.089. . . moles:

(1.50) (x) = (0.089276229) (0.08206) (298.0)

x = 1.46 L

Problem #17: A mixture of CO_{2} 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 CO_{2} is completely removed by absorption with NaOH(s) the pressure in the container is 0.250 atm. How many grams of CO_{2} and how many grams of Kr were initially present?

Solution:

1) Calculate the mole fractions of CO_{2} and Kr:

CO_{2}: 0.458 atm / 0.708 atm = 0.6469

Kr: 0.3531

2) Change these to “gram fractions:”

CO_{2}: 0.6469 x 44.0 = 28.4636

Kr: 0.3531 x 83.8 = 29.59

CO_{2}: 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:

CO_{2}: 35.0 x 0.49 = 17.15 g

Kr: 17.85 g

Comment: Based on this ratio (0.458/0.250) the CO_{2}: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 CO_{2} 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 = nRT

PV/T = nR

Since the right-hand side is constant, the answer is B.

Problem #19: Our atmosphere is a mixture of gases (roughly 79% N_{2}, 20% O_{2} and 1%Ar).

(a) What is the partial pressure (in atm) of each gas in the atmosphere?

(b) A mixture of He and O_{2} 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 O_{2}, if the partial pressure of O_{2} 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:

N_{2}: 79 g / 28.0 g mol¯1 = 2.82 mol

O_{2}: 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/O_{2} mixture:

0.180 / 8 = 0.0225 mol of O_{2}

7.820 / 8 = 0.9775 mol of He

2) Convert each to grams:

O_{2}: 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 O_{2} in the mixture:

0.72 g / 4.63 g = 0.1555 = 15.55%

Problem #20: 600.0 mL of a mixture of O_{2} and O_{3} 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 = 0.024536

multiply 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 N_{2} 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 N_{2} 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 Pressure atm = 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 = 0.0833

where m = mass of Mg and z = mass of Zn

4) Set up second equation:

(m / 24.305) + (z / 65.38) = 0.0023787

m / 24.305 = moles of magnesium

z / 65.38 = moles of zinc

The 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) + H_{2}SO_{4} (aq) —> H_{2} (g) + ZnSO_{4} (aq)

Mg (s) + H_{2}SO_{4} (aq) —> H_{2} (g) + MgSO_{4} (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 g

To three sig figs:

m = 0.0427 g

z = 0.0406 g

If you use the on-line app, make sure to use these values for the second equation:

m / 24.305 = 0.0411438m

z / 65.38 = 0.015295z

As an additional exercise, you may wish to solve the equation system by hand.

Problem #23: A mixture of oxygen and helium is 92.3% by mass oxygen. It is collected at atmospheric pressure (745 torr). What is the partial pressure of oxygen in this mixture?

Solution:

Let’s assume a big, big volume such that this volume holds 100 g of the mixture at 745 torr and whatever the temperature is.

92.3 g of the 100 g is O_{2} and 7.7 g is He.

Compute moles of each because pressure is proportional to the number of particles:

O_{2} —> 92.3 g / 32.0 g/mol = 2.884375 mol

He —> 7.7 g / 4.0026 g/mol = 1.92375 mol

We now will calculate what we need to solve this equation: partial pressure = mole fraction x total pressure.

Get total moles:

2.884375 + 1.92375 = 4.808125 mol

Get mole fraction of O_{2}:

2.884375 / 4.808125 = 0.599896 = 0.600

Get partial pressure of O_{2}:

0.600 times 745 torr = 447 torr

Problem #24: If 20 percent of 100 mL sample of oxygen is converted into ozone, what would be its final volume?

Solution:

Let us split the volume into two: one of 20 mL and one of 80 mL. Leave the 80 mL untouched and allow all the 20 mL to react.

3O_{2} —> 2O_{3}

We will keep the pressure and temperature constant.

We will use Gay-Lussac’s Law of Combining Volumes:

three volumes of O_{2} will react to form two volumes of O_{3}

3 is to 2 as 20 is to x

x = 13.3 mL

Combine the two volumes:

80 + 13.3 = 93.3 mL

Problem #25: Low-pressure gauges in research laboratories are occasionally calibrated in inches of water. The density of mercury at 15 °C is 13.5 g/cm3 and the density of water at that temperature is 1.00 g/cm3. What is the pressure (in torr) inside a gas cylinder that reads 1.40 in. H_{2}O at 15 °C?

Solution:

Mercury is 13.5 times denser than water, so it takes 13.5 times as much water to indicate the same pressure as compared to mercury.

Another way to express this is that the pressure times the density of the indicating liquid (water or mercury) is a constant.

(1.40 inch) (1.00 g/cm3) = (x) (13.5 g/cm3)

x = 0.1037 inch of mercury

The pressure unit torr is measured in millimeters of mercury.

0.1037 inch times 25.4 mm/inch = 2.63 torr (rounded to three sig figs)

Based on theories of the big bang nucleosynthesis, physicists in the mid-1960s calculated that roughly 1/4 of mass was converted into helium in the beginning, while the rest remained as hydrogen. This would be consistent with the earlier measurements of 20-30% helium abundance if most of the observed helium were present, from the big bang, even before stars began producing more. In the early 1970s, spectroscopic studies in other galaxies have confirmed that the majority of the observed helium did exist before any star formation. The figure shows the helium abundance of many galaxies having various oxygen abundances. Oxygen abundance indicates in general how much nucleosynthesis has taken place in stars because stars produce “heavy” elements (like oxygen, nitrogen, carbon, and helium) from hydrogen through nuclear fusion. If all observed helium was created in stars like all oxygen was, we would expect to find no helium in galaxies that have no oxygen because these galaxies must have formed before any stars created heavy elements. Yet, as the plot shows, the abundance of helium decreases very little as the abundance of oxygen approaches zero. The galaxies must have formed with an initial composition of about 24% helium. This observational result supports the theory that there must have been a big bang in the beginning which converted about 1/4 of mass into helium through nucleosynthesis.