CHEM 1105 GASES

1. Properties of Gases Gases, like liquids, are fluids, that is, they can flow. However, in gases, unlike liquids, the particles (atoms or molecules) are far apart. This leads to several special properties of gases. (a) All gases are miscible. (b) Gases are compressible (their volumes decrease as the pressure increases). (c) Intermolecular forces between gas particles are negligible. (d) Gases expand to fill the container. (e) Collision of gas particles with the walls of the container exerts a pressure. (f) The pressure of a gas in a closed container increases as the increases. (g) The pressure of a gas in a closed container increases as the number of molecules of the gas increases.

Because of the properties discussed above, the pressure and temperature of a gas, as well as its , must be specified.

2. Pressure F Pressure is defined as force per unit area (P = ). Force is mass x acceleration and in SI A F = kg·m·s-2 = Newtons (N). Hence P = kg·m-1·s-2 = Pascals (Pa). Other units used by chemists are mmHg (or torr) (pressure exerted by a 1 mm column of Hg) and atmospheres (1 atm = 760 mmHg). Volume of a cylinder of liquid, V = Ah (h = height of cylinder) Mass of a cylinder of liquid, M = Vd (d = density of liquid) = Ahd Force exerted by gravity on cylinder of liquid = Mg (g = acceleration due to gravity) = Ahdg Pressure = F/A = Ahdg/A = hdg

-2 3 3 g = 9.807 m·s ; dHg = 13.6 g/cm = 13600 kg/m ; 760 mm = 0.760 m Therefore, a 760 mm column of Hg exerts a pressure of 9.807 x 13600 x 0.760 = 1.01 x 105 Pa Therefore, 1 atm = 1.01 x 105 Pa = 101 kPa

Because the density of water is about 1.00 g/mL, it would take a column of water of about 10 m to exert a pressure of 1 atm. (0.760 x 13.6 = 10.3)

N.B.: Barometers are used for measuring atmospheric pressure and manometers are used for measuring the pressure of a gas in a container.

3. Temperature The temperature scale used in gas measurements must be an absolute scale. Chemists use the scale (K = °C + 273.15).

(a) Convert 0 K to °F (Ans. -459.67°F)

(b) Alcohol boils at 78.5°C and freezes at -117.3°C. On the booze scale, the of alcohol is 100°B and the freezing point of alcohol is 0°B. Convert 25.0°C to °B. (Ans. 72.7°B) -2-

4. Ideal Gas Law Because the forces between gas molecules are negligibly small, all gases behave approximately the same. Except under special conditions (to be discussed later), all gases obey the Ideal Gas Law PV = nRT (P = pressure, V = volume, n = moles, R = Universal Gas Constant, T = absolute temperature). Experimentally, it was found that the volume of 1 mole of any gas at STP (standard temperature and pressure: 0°C, 273 K, and 1 atm) is about 22.4 L. Substituting in the equation, 1 atm x 22.4 L = 1 mole x R x 273 K. Hence R = 0.0821 L·atm/K·mole. This value of R will usually be used and will always be provided. Using the pressure as 760 mmHg, 760 mmHg x 22.4 L = 1 mole x R x 273 K and R = 62.4 L·mmHg/K·mole. In SI units, 1 atm pressure = 101.3 x 103 Pascals and volume is in m3. Since L = dm3, 1000 L = 1 m3 and 22.4 L = 0.0224 m3. Hence R in SI units can be derived from 101.3 x 103 Pa x 0.0224 m3 = 1 mole x R x 273 K, giving R = 8.31 J/K·mole. It can easily be shown that using pressure in kPa and volume in L, R = 8.31 L·kPa/mole·K

The derivation of R and P in SI units will not be examined!

PV = nRT is used to calculate V if P, n and T are known, to calculate n if P, V and T are known, etc. Since n = mass/molar mass, and density = mass/volume, the equation can also be used to calculate molar mass and density of a gas. Examples will be given later.

Historically, the Ideal Gas Law was obtained by the combination of several experimentally determined gas laws. As shown below, these laws can be seen to be special cases of the ideal gas law.

1. If n and T are kept constant (a fixed amount of gas is kept at constant temperature), PV = nRT becomes PV = constant. Hence, volume is inversely proportional to pressure (Boyle’s Law). This leads to P1V1 = P2V2.

(a) A certain mass of gas at constant temperature has a volume of 45.0 mL at 755 mmHg. What will the volume be at 375 mmHg? (Ans. 90.6 mL)

(b) A certain mass of gas at constant temperature has a volume of 326 mL at 0.950 atm. At what pressure will the volume be 125 mL? (Ans. 2.48 atm)

2. If n and P are kept constant (a fixed amount of gas is kept at constant pressure), PV = nRT becomes V = constant x T. Hence, volume is directly proportional to the absolute temperature

V1 V2 (Charles’ Law). This leads to = . T1 T2

(c) A certain mass of gas at constant pressure has a volume of 22.0 mL at 25.0°C. What will the volume be at 75.0°C? (Ans. 25.7 mL)

(d) To what temperature (in °C) must a gas at 20.0°C be raised to double its volume? (Ans. 313°C) -3-

PV 3. If n is kept constant (a fixed amount of gas) PV = nRT becomes = constant. This is a T

P1V1 P2V2 combination of Boyle’s Law and Charles’ Law and leads to = . T1 T2

(e) A certain mass of gas at 32.5°C and 743 torr has a volume of 268 mL. What will the volume be at 68.5°C and 0.500 atm? (Ans. 586 mL)

4. If n and V are kept constant (a fixed amount of gas is kept at constant volume), PV = nRT becomes P = constant x T (Amonton’s Law).

5. If T and P are kept constant (a gas at constant temperature and pressure), PV = nRT becomes V = constant x n.

Problems using PV = nRT

° (f) Calculate the volume occupied by 2.68 g of N2O at 19.0 C and 755 mmHg. (Ans. 1.47 L)

(g) Calculate the molar mass of a gas if 1.50 g of the gas in a 670 mL container has a pressure of 0.950 atm at 27.0°C. (Ans. 58.0 g/mole)

° (h) Calculate the density of CH3NCO at 31 C and 770 torr. (Ans. 2.31 g/L) ° (i) Calculate the temperature (in C) of 0.125 g of C2H6 in a 250.0 mL container at a pressure of 255 mmHg. (Ans. -27.8°C)

(j) How many N2 molecules are left in a 1.00 L vessel containing N2 which has been evacuated to a pressure of 1.00 x 10-2 torr at 25°C? (Ans. 3.24 x 1017)

5. Gas Stoichiometry PV = nRT allows the volume of a gas at known temperature and pressure to be converted into moles.

(a) What volume of ammonia, measured at STP, would be obtained from the reaction of 4.00 g of Mg3N2 with an excess of water. The equation is given below. (Ans. 1.78 L) H Mg3N2(s) + 6H2O(l) 3Mg(OH)2(s) + 2NH3(g)

(b) What mass of Mg3N2 must react with excess water to produce 868 mL of NH3(g) at 765 torr and 23.5°C? Assume a 100% yield. (Ans. 1.81 g)

(c) Air bags in cars contain sodium azide which is decomposed by shock as shown:

H 2NaN3(s) 2Na(s) + 3N2(g)

Calculate the pressure generated in an evacuated 5.00 L container when 10.0 g of sodium azide decompose at 25.0°C. (Ans. 1.13 atm) -4-

(d) NH4NO2 decomposes on heating to form N2 and water as shown below: H NH4NO2(s) N2(g) + 2H2O(g)

A 2.00 g sample of NH4NO2 was placed in an evacuated 800.0 mL container and completely decomposed by heating to 155°C. Calculate the pressure when the contents of the container were at (1) 155°C (Ans. 4.12 atm), and (2) cooled to 25°C. (Ans. 0.956 atm)

6. Dalton’s Law of Partial Pressures Dalton’s Law of partial pressures states that the pressure of a mixture of gases is equal to the sum of their partial pressures. The of a gas in a mixture is the pressure it would exert if it alone occupied the whole volume.

Ptotal = pA + pB The law can be verified, as shown below. RT RT RT RT P = n x = (n + n ) x = n x + n x total total V A B V A V B V

RT RT But, n x = p and n x = p , therefore P = p + p A V A B V B total A B

Gases that do not react with water or dissolve in water are often collected over water (see p. 202). Dalton’s Law is useful in dealing with gases collected over water because the measured pressure is the sum of the partial pressure of the gas and the partial pressure of the water vapour (the vapour pressure of water). The vapour pressure for water depends only on the temperature of the water and is known for a wide range of .

Pgas + VPwater = measured (barometric) pressure pgas = measured pressure - VPwater

(a) A 75.0 mL sample of a gas was collected over water at 22°C and a barometric pressure of 0.987 atm. What volume would the gas occupy if dry and at 100.0°C and 1.000 atm? The vapour pressure of water at 22°C is 19.8 torr. (Ans. 91.1 mL)

(b) Calculate the volume of hydrogen gas, collected over water at 29°C and a barometric pressure of 768 mmHg, which would be obtained from the reaction of 0.300 g of zinc with excess hydrochloric acid. The vapour pressure of water at 29°C is 30.0 mmHg. The reaction is shown below. (Ans. 117 mL)

H Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

(c) When 0.450 g of an unknown metal M was reacted with an excess of hydrochloric acid and the hydrogen evolved was collected over water, 595 mL of hydrogen gas at 775.2 torr and 22°C were obtained. The vapour pressure of water at 22°C is 19.8 torr. Calculate the atomic weight of the metal and hence identify the metal. The reaction is shown below. (Ans. Al)

H 2M(s) + 6HCl(aq) 2MCl3(aq) + 3H2(g) -5-

Using the vapour pressure of water at a given temperature, the mass of water vapour in a certain volume above liquid water can be calculated using PV = nRT.

(d) Calculate the mass of water vapour per litre above water in a closed container at 22°C. (Ans. 0.0194 g or 19.4 mg)

From the problem above, it can be seen that if a 1.00 L container at 22°C had a mass of water equal to or less than 19.4 mg, all of the water would vapourize.

Partial Pressure vs Mole Fraction RT RT p = n x P = n x A A V total total V

pA nA nA pA x P Therefore, = n and = n total Ptotal total total

nA = mole fraction of A = X (N.B.: X + X = 1 for a mixture of 2 gases) ntotal A A B

pA hence, pA = XA x Ptotal and XA = Ptotal

(e) A mixture of 2.5 moles of CO and 3.5 moles of NO has a pressure of 0.84 atm. Calculate the partial pressures of CO and NO. (Ans. pCO = 0.35 atm; pNO = 0.49 atm)

(f) A mixture of 7.00 g of CO and 5.50 g of CO2 exerts a pressure of 0.900 atm. Calculate the partial pressure of each gas. (Ans. pCO = 0.600 atm; pCO2 = 0.300 atm)

(g) In a mixture of N2 and C2H6 gases, the partial pressures of N2 and C2H6 were 0.300 atm and 0.450 atm, respectively. (1) Calculate the mole fractions of N2 and C2H6. (Ans. 0.400 and 0.600) ° (2) Calculate ntotal if the volume is 20.4 L at 100 C. (Ans. 0.500) (3) Calculate the mass (in grams) of C2H6. (Ans. 9.0 g)

(h) A 400 mL container of He at 1.00 atm was connected to a 100 mL container of Ar at 2.00 atm by a tube of negligible volume with a closed stopcock. The stopcock was then opened, allowing the gases to mix. Calculate (1) the final pressure in the system (Ans. 1.20 atm), and (2) the mole fraction of Ar in the mixture (Ans. 0.333) (Hint: for each gas, P1V1 = P2V2)

(i) A 600 mL container of N2 at 1.5 atm was connected to a container of CO2 of unknown volume at 2.5 atm by a tube of negligible volume with a closed stopcock. The stopcock was then opened, allowing the gases to mix. The final pressure in the system was found to be 2.2 atm. (1) Calculate the initial volume of CO2. (Ans. 1400 mL) ° (2) If the temperature was 37 C, calculate the mass of CO2. (Ans. 6.1 g)