Warmup

A 40.0m3 weather balloon at standard temperature encounters a sudden icy cool breeze during its journey into the Alaskan sky. If the weather balloon decreases to 37.9 m3, calculate the temperature of the icy cool breeze. Assume pressure is held constant at 610 mmHg.

The Combined Gas Law and Dalton’s Law of Partial

PressuresYou have all the sample problems

in your packet

P1V1 = P2V2

T1 T2

(101.3 kPa)(V1) = (99.0 kPa) (466 ml )

(273K) (-4.6°C+273)(101.3kPa)(V1)(268.4°K)=(99.0 kPa)(466 ml)(273K)

(101.3 kPa)(268.4K) (101.3 kPa)(268.4K)

V1 = 460 ml

Ex1. A half-full water bottle (at STP) travels by car and is put in a freezer high up in the mountains, changing the temperature to -4.6° C. The volume changes to 466 ml and the pressure decreases to 99.0 kPa. What was the original volume of the gas inside the water bottle?

Ex 2: A helium balloon at standard, constant temperature is placed in a vacuum, decreasing the pressure to 0.843 atm. The volume expands from 1.78 L to 4.98 L. What was the pressure of the balloon in the beginning?

Use Boyle’s Law Use Combined Gas Law

same answer…. I swear!

P1V1 = P2V2

T1 T2

P1(1.78 L)= (0.843 atm) (4.98 L)

(273K) (273K)

P1(1.78 L) (273K) =(0.843 atm)(4.98 L)(273K)

(1.78 L) (273K) (1.78 L) (273K)

P1 = 2.36 atm

Ex 3: A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.°C?

Initial After

P1 = P2 =

V1 = V2 =

T1 = T2 =

P1T1

P2 T2

1.08 atm

0.855 atm

50.0 L ?

25°C298 K

10.°C283 K

2

22

1

11

TVP

TVP

V2 =

atm 0.855 X K 298

K283 X L 50.0 X atm 1.08

V2 = 60. L

V1

V2

21

211

PTTVP

He CO2 H2O

0.7 atm 0.8 atm 0.8 atm 2.3 atm

Dalton’s Law of Partial Pressuresthe total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases

Ptotal = P1 + P2 + P3 + …

Ex 1: A container holds three gases: oxygen, carbon dioxide, and helium. The partial pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is the total pressure inside the container?

Ptotal = P1 + P2 + P3 + …

Ptotal = PO2 + + PCO2 + PHe

Ptotal = 2.00 atm + 3.00 atm + 4.00 atm

Ptotal = 9.00 atm

Gas Collected Over Water:• you need to account for the additional

vapor pressure :

Ptotal = Pgas + PH2ONote: most of the time, the total pressure is equal

to the pressure inside the room!

Ex 2: Helium gas is collected over water at 25 °C. What is the partial pressure of the helium, given that the barometric pressure is 750.0 mm Hg?

Ptotal = PHe + PH2O

750.0 mm Hg = PHe + 23.8 mm Hg

PHe = 726.2 mm Hg

Ex 3: A student has stored 100.0 mL of neon gas over water on a day when the temperature is 28.0°C. If the barometer in the room reads 743.3 mm Hg, what is the pressure of the neon in its container?

Ptotal = PNe + PH2O

743.3 mmHg = PNe + 28.3 mm Hg

PNe = 715.0 mm Hg

Ex 4: A container has two gases, helium and argon. Suppose that the mixture contains 30% helium by volume. Calculate the partial pressure of helium AND argon if the total pressure of the container is 4.00 atm.

Total pressure = 4.00 atm.30% of 4.00 is 1.2 atm (0.30 X 4.00 atm = 1.2 atm)70 % of 4.00 is 2.8 atm (0.70 X 4.00 atm = 2.8 atm)

PHe = 1.2 atm AND PAr = 2.8 atm

Double Check: Ptotal = PHe + + PAr

Ptotal = 1.2 atm + 2.8 atm = 4.00 atm

Ex 5: A certain mass of oxygen was collected over water when potassium chlorate was decomposed by heating. The volume of the oxygen sample collected was 720 mL at 25°C and an atmospheric pressure of 750 mmHg. What would the volume of the oxygen be at STP?

P1V1 = P2V2

T1 T2

P1(720ml) = (760mmHg)V2

(273 + 25⁰C) (273K)

Ptotal = PO2 + PH2O

750 mmHg = PO2 + 23.8 mmHg

PO2 = 726.2 mmHg

Answer: 630 mL

(726.2 mmHg)(720ml) = (760mmHg)V2

(298K) (273K)

Mystery Game:The Gas Problem Strikes BACK

Directions:

You will be given three TIMED gas law problems. (Exactly 3 minutes to do each

problem…so work together). Your job is to find an answer to each mystery problem. The sum of all three mystery problems, ignoring

sig fig rules, is: 757.15

The volume of a gas at 27.0°C and 0.200 atm is 80.0 mL. What volume will the same gas sample occupy at standard temperature and standard pressure?

What pressure(atm) is required to reduce 60.0 mL of a gas at STP to 10.0 mL at a temperature of 25.0°C?

A 250. mL sample of oxygen gas is collected over water at 25.0 °C and 760 mmHg. What is the pressure of the oxygen gas by itself?

add your answers to all 3 mystery problems and round to 3 sig figs!

Mystery Problem

#1

Mystery Problem

#2

Mystery Problem

#3

ANSWER (no units)+ + =

14.6 mL

+6.55

atm+ 736

mm Hg= 757.15