The Gas Laws-Part I The Gaseous State

The States of Matter

The Distinction of Gases from Liquids and Solids 1. Gas volume changes greatly with pressure. 2. Gas volume changes greatly with temperature. 3. Gases have relatively low viscosity. 4. Most gases have low densities. 5. Gases are miscible.

The Pressure of a Gas Gas pressure is a result of the constant movement of the gas molecules and their collisions with the surfaces around them. The pressure of a gas depends on 1) number of gas particles 2) volume of the container 3) speed of the gas particles

A Mercury Barometer We measure air pressure with a barometer. Column of mercury supported by air pressure gravity Force of the air on the surface of the mercury counter balances the force of gravity on the column of mercury.

Pressure explains how you can drink from a straw. a) When a straw is put into a glass of orange soda, the pressure inside and outside of the straw are the same, so the liquid levels inside and outside of the straw are the same. b) When a person sucks on the straw, the pressure inside the straw is lowered. The greater pressure on the surface of the liquid outside of the straw pushes the liquid up the straw.

A drinking straw s limitations. Even if you formed a perfect vacuum, atmospheric pressure could only push orange soda to a total height of about 10 m. This is because a column of water 10 m high exerts the same pressure (14.7 lb/in 2 ) as the gas molecules in our atmosphere.

Common Units of Pressure

Conversion Between Pressure Units psi atm mmhg 1 atm = 14.7 psi 1 atm = 760 mmhg 1.00 atm 14.7 psi 14.7 psi 1.00 atm 1.00 atm 760. mm Hg 760 mm Hg 1.00 atm 1 torr = 1 mmhg

How do scientists look at data???

a = independent variable b = dependent variable a b (a)/(b) (a)(b) 1 2 0.5 2 3 6 0.5 18 5 10 0.5 50 7 14 0.5 98 (a)/(b) = constant (a)/(b) = k (a) = (k)(b); a b (b) = (k)(a); b a a direct proportion 9 18 0.5 162

a = independent variable b = dependent variable a b (a)/(b) (a)(b) 1 2 0.5 2 3 0.67 4.5 2 5 0.4 12.5 2 7 0.28 25 2 9 0.22 41 2 (a)(b) = constant (a)(b) = k (a) = (k)/(b); a 1/b (b) = (k)/(a); b 1/a an inverse proportion

Boyle s Law Robert Boyle (1627 1691) Pressure of a gas is inversely proportional to its volume (constant T and amount of gas) As P increases, V decreases by the same factor.

Boyle s Experiment

Boyle s Law Pressure of a gas is inversely proportional to its volume (constant T and amount of gas). As P increases, V decreases by the same factor. P= 1.0 atm P= 2.0 atm increase pressure V= 1.0 L decrease pressure V= 0.5 L

Boyle s Law V vs P V vs 1/P 20 20 Volume (L) 15 10 5 Volume (L) 15 10 5 0 0 200 400 600 800 Pressure (mm Hg) 0 0.0000 0.0055 0.0110 0.0165 0.0220 1/P V 1/P V = k/p P x V = constant P1 x V1 = P2 x V2

1. A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm? V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm, V2=? P1 V1 = P2 V2 V1, P1, P2 V2 V2 = (P1)(V1) = (P2) (4.52 atm)(7.25 L) (1.21 atm) = 27.1 L

2. A balloon has a volume of 1.35 x 10 3 ml at 782 torr. It is put in a bell jar and the pressure is reduced by applying a vacuum pump. If the volume of the balloon is now 2.78 x 10 3 ml, what is the pressure inside the bell jar? V1 = 1350 ml, P1 = 782 torr, V2 = 2780 ml P2 =? V1, P1, V2 P2 P1 V1 = P2 V2 P2 = (P1)(V1) = (V2) (782 torr)(1350 ml) (2780 ml) = 380 torr

Charles s Law Jacques Charles (1746 1823) Volume is directly proportional to temperature (constant P and amount of gas) As T increases, V also increases Kelvin T = Celsius T + 273

Charles s Law At constant n and P, the volume of a gas increases proportionately as its absolute temperature increases. If the absolute temperature is doubled, the volume is doubled. P= 1.0 atm P= 1.0 atm increase temperature V= 0.5 L decrease temperature V= 1.0 L T = 200 K T = 400 K

Charles s Law V vs T(K) 1.40 1.05 Volume (L) 0.70 0.35 0.00 0 100 200 300 400 Temperature (K) V T V = kt

If the lines are extrapolated back to a volume of 0, they all show the same temperature, 273.15 C, called absolute zero.

3. A gas has a volume of 2.57 L at 0.00 C. What was the temperature when the gas had a volume of 2.80 L? V2 =2.57 L, T2 = 0.00 C, V1 = 2.80 L T1 =? V2, T2, V1 T1 T(K) = t(ºc) + 273.15 V1 T1 = V2 T2 T2 = 0.0 + 273.15 = 273.15 K T1 = (T2)(V1) = (V2) (273.15 K)(2.80 L) (2.57 L) = 297.6 K = 24.4 ºC

4. The temperature inside a balloon is raised from 25.0 C to 250.0 C. If the volume of cold air was 10.0 L, what is the volume of hot air? V1 =10.0 L, t1 = 25.0 C L, t2 = 250.0 C V2 =? V1, T1, T2 V2 T(K) = t(ºc) + 273.15 V1 T1 = V2 T2 T1 = 25.0 + 273.15 = 298.2 K T2 = 250.0 + 273.15 = 523.2 K V2 = (T2)(V1) = (T1) (523.2 K)(10.0 L) (298.2 K) = 17.5 L

Amonton s Law Guillaume Amontons (1663 1705) Pressure is directly proportional to the absolute temperature of gas molecules (constant n and V) Higher temperature = Higher pressure P T P = kt P1 T1 = P2 T2

Amonton s Law P vs T 1200 Pressure (torr) 1000 800 600 400 200 0 0 80 160 240 320 400 Temperature (K) P T P = kt P1 T1 = P2 T2

Combined Gas Law Volume is inversely proportional to pressure and directly proportional to temperature (for a constant amount of gas) As P increases, V decreases (T, constant) As T increases, V increases (P, constant) As T increases, P increases (V, constant) V 1 P V T V T P PV T = K P 1 V 1 P = 2 V 2 T 1 T 2

Avogadro s Law Equal volumes of gases, under the same conditions of temperature and pressure, contain the same number of molecules.

Avogadro s Law Amedeo Avogadro (1776 1856) Volume is directly proportional to the number of gas molecules (constant P and T) More gas molecules = larger volume Equal volumes of gases contain equal numbers of molecules. The identity of the gas doesn t matter!! V n V = kn

Avogadro s Law At constant T and P, the volume of a gas increases proportionately as its molar amount increases. If the molar amount is doubled, the volume is doubled. P= 1.0 atm P= 1.0 atm add gas V= 22.4 L remove gas V= 44.8 L n = 1.0 mol n = 2.0 mol

Avogadro s Law V vs n 400 Volume(L) 300 200 100 0 0 2 4 6 8 10 number of moles (n) V n V = kn

5. If 1.00 mole of a gas occupies 22.4 L and the temperature and pressure are not changed, what volume would 0.750 moles occupy? V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol V2 =? V1, n1, n2 V2 V1 n1 = V2 n2 V2 = (V1)(n2) (n1) = (22.4 L)(0.750 mol) (1.00 mol) = 16.8 L

Standard Conditions Because the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy we call these standard conditions - STP Standard pressure = 1 atm Standard temperature = 273 K, 0 C

V 1 P V T V n V nt P V =(k) nt P PV k= = R = nt (1.00 atm)(22.414 L) (1.00 mole)(273.15 K) R = 0.0821(atm)(L)/(mol)(K)

Ideal Gas Law By combining the gas laws we can write a general equation: R is called the gas constant. The value of R depends on the units of P and V. We will use 0.0821 and convert P to atm and V to L. atm L mol K

Boyle s Law PV = k Charles s Law V/T = k n,r,t constant n,r,p constant Ideal Gas Law PV = nrt n,r,v constant T,R,P constant Amonton s Law P/T = k Avogadro s Law V/n = k

Ideal Gas Law PV = nrt n and R constant P1V1 T1 = P2V2 T2 Combined Gas Law PV/T = k

6. Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 10 4 mmhg and 23 C. mso2 = 637 g, P = 6.08 x 10 4 mmhg, t = 23 C, V =? g n P, n, T, R V 1.00 mol 64.07 g SO2 1.00 mol SO2 637 g SO2 x = 9.942 mol SO2 64.07 g SO2 1.00 atm 6.08 x 10 4 mmhg x = 80.0 atm 760 mm Hg V = (n)(r)(t) P K = 273+ºC = 273+(-23) = 250 atm L (9.942 mol)(0.0821 mol K )(250 K) 80.0 atm = 2.55L

7. A gas occupies 10.0 L at 44.1 psi and 27 C. What volume will it occupy at standard conditions? V1 = 10.0L, P1 = 44.1 psi, t1 = 27 C, P2 = 1.00 atm, t2 = 0 C, V2 =? P1, V1, T1, R n n = PV RT P2, n, T2, R V2 V = nrt P 44.1 psi x 1.00 atm = 3.00 atm 14.7 psi n = PV RT = (3.00 atm)(10.0 L) atm L (0.0821 mol K )(300. K) T1(K) = 27ºC + 273.15 = 300. K = 1.218 mol = 1.22 mol

7. A gas occupies 10.0 L at 44.1 psi and 27 C. What volume will it occupy at standard conditions? P1, V1, T1, R n n = PV RT P2, n, T2, R V2 V = nrt P V = atm L (1.22 mol)(.0821 )(273 K) mol K 1.00 atm n =1.22 mol = 27.3 L

7A. A gas occupies 10.0 L at 44.1 psi and 27 C. What volume will it occupy at standard conditions? This is a comparison of two sets of conditions; you can use the combined gas law. P 1 = 3.00 atm V 1 = 10.0 L T 1 = 300 K P 2 = 1.00 atm T 2 = 273 K V 2 =? V2 = P1V1 T1 V2 = = P2V2 T2 P1V1T2 T1P2 (3.00 atm)(10.1l)(273 K) (300 K)(1.00 atm) V2 = 27.3 L