Gases

The Combined Gas Law

Volume and Moles (Avogadro’s Law)

Partial Pressures (Dalton’s Law)

Combined Gas Law

P1V1 = P2V2

T1 T2

Problem 1

Solve the combined gas laws for T2.

Solution for Problem 1

Solve the combined gas law for T2.

(Hint: cross-multiply first.)

P1V1 = P2V2

T1 T2

P1V1T2 = P2V2T1

T2 = P2V2T1

P1V1

Combined Gas Law Problem

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Data Table

Set up Data Table

P1 = 0.800 atm V1 = 0.180 L T1 = 302 K

P2 = 3.20 atm V2= 90.0 mL T2 = ????

Solution

Solve for T2

Enter data

T2 = 302 K x atm x mL = K

atm mL

T2 = K - 273 = °C

Calculation

Solve for T2

T2 = 302 K x 3.20 atm x 90.0 mL = 604 K

0.800 atm 180.0 mL

T2 = 604 K - 273 = 331 °C

Try this problem

A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

Solution

T1 = 308 K T2 = ?

V1 = 675 mL V2 = 0.315 L = 315 mL

P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg

T2 = 308 K x 802 mm Hg x 315 mL

646 mm Hg 675 mL

P inc, T inc V dec, T dec

Solution

T1 = 308 K T2 = ?

V1 = 675 mL V2 = 0.315 L = 315 mL

P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg

T2 = 308 K x 802 mm Hg x 315 mL

646 mm Hg 675 mL

P inc, T inc V dec, T dec

= 178 K - 273 = - 95°C

Volume and Moles

How does adding more molecules of a gas change the volume of the air in a tire?

If a tire has a leak, how does the loss of air (gas) molecules change the volume?

Answer these questions

True (1) or False(2)

1.___The P exerted by a gas at constant V is not affected by the T of the gas.

2.___ At constant P, the V of a gas is directly proportional to the absolute T

3.___ At constant T, doubling the P will cause the

V of the gas sample to decrease to one-half its

original V.

Solution

True (1) or False(2)

1. (2)The P exerted by a gas at constant V is not affected by the T of the gas.

2. (1) At constant P, the V of a gas is directly proportional to the absolute T

3. (1) At constant T, doubling the P will cause the

V of the gas sample to decrease to one-half its

original V.

Avogadro’s Law

When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas

V1 = V2

n1 n2

initial final

STP

The volumes of gases can be compared when they have the same temperature and pressure. If the temperature and pressure are:

Standard temperature 0°C or 273 K

Standard pressure 1 atm (760 mm Hg)

then, it is represented by STP.

Try this problem

A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

P1 = V1 = T1 = K

P2 = V2 = ?? T2 = K

V2 = 15 L x atm x K = 6.8 L

atm K

Solution

P1 = 1.0 atm V1 = 15 L T1 = 273 K

P2 = 2.0 atm V2 = ?? T2 = 248 K

V2 = 15 L x 1.0 atm x 248 K = 6.8 L

2.0 atm 273 K

Molar Volume

At STP

4.0 g He 16.0 g CH4 44.0 g CO2

1 mole 1 mole 1mole (STP) (STP) (STP)

V = 22.4 L V = 22.4 L V = 22.4 L

Molar Volume Factor

1 mole of a gas at STP = 22.4 L

22.4 L and 1 mole

1 mole 22.4 L

Try these problems

A.What is the volume at STP of 4.00 g of CH4?

1) 5.60 L 2) 11.2 L 3) 44.8 L

B. How many grams of He are present in 8.0 L

of gas at STP?

1) 25.6 g 2) 0.357 g 3) 1.43 g

Solution

A.What is the volume at STP of 4.00 g of CH4?

4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L

16.0 g CH4 1 mole CH4

B. How many grams of He are present in 8.0 L of gas at STP?

8.00 L x 1 mole He x 4.00 g He = 1.43 g He

22.4 He 1 mole He

Daltons’ Law of Partial Pressures

Partial Pressure

Pressure each gas in a mixture would exert if it were the only gas in the container

Dalton's Law of Partial Pressures

The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture.

PT = P1 + P2 + P3 + .....

Gases in the Air

The % of gases in air Partial pressure (STP)

78.08% N2 593.4 mmHg

20.95% O2 159.2 mmHg

0.94% Ar 7.1 mmHg

0.03% CO2 0.2 mmHg

PAIR = PN + PO + PAr + PCO = 760 mmHg 2 2 2

Total Pressure 760 mm Hg

Dalton’s Law Check

A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?

1) 35.6 2) 156 3) 760

B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?

1) 557 2) 9.14 3) 0.109

Solution

A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in the air?

2) 156

B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?

1) 557

Partial Pressures

The total pressure of a gas mixture depends

on the total number of gas particles, not on

the types of particles.

P = 1.00 atm P = 1.00 atm

0.5 mole O2

+ 0.3 mole He+ 0.2 mole Ar

1 mole H2