Gas Laws

As T increases

V increases

Charles’s LawV1 x T2 = V2 x T1

At constantPressure andAmount of gas

A sample of carbon monoxide gas occupies 3.20 L at 125°C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?

V1 = 3.20 L

T1 = 125°C + 273 = 398 K

V2 = 1.54 L

T2 = ?

T2 = V2 x T1

V1

1.54 L x 398 K3.20 L

= = 192 K

V1 x T2 = V2 x T1

*Temperature must always be in Kelvin*

A sample of carbon dioxide gas occupies 6.91 L at 112°C. What volume will the gas occupy when the temperature is raised to 179°C if the pressure remains constant?

V1 = 6.91 L

T1 = 112°C + 273 = 385 K

V2 = ?

T2 = 179°C + 273 = 452 K

V2 = V1 x T2

T1

6.91 L x 452 K385 K

= = 8.11 L

V1 x T2 = V2 x T1

*Temperature must always be in Kelvin*

Combined Gas Law

If the initial volume of a gas was 405 mL at 90.0°C and 0.581 atm, what would the volume be at STP?

V2 = P1V1T2 P2T1

T1 = 90.0°C + 273 = 363 K

P1 = 0.581 atm

V2 = (1.00 atm)(363K)

(0.581 atm)(405mL)(273K)

V2 = 177 mL

P1V1T2 = P2V2T1

V1 = 405 mL

T2 = 0 0C + 273 = 273 K

P2 = 1.00 atm

V2 = ?

Combined Gas Law

If the initial volume of SO2 was 236 L at 75.0°C and 1.25 atm, what would the pressure be when the

volume is raised to 652 L while the temperature is decreased to 25.0°C?P2 = P1V1T2

V2T1

T1 = 75.0°C + 273 = 348 K

P1 = 1.25 atm

P2 = (652 L)(348K)

(1.25 atm)(236 L)(298K)

P2 = 0.387 atm

P1V1T2 = P2V2T1

V1 = 236 L

T2 = 25.00C + 273 = 298 KV2 = 652 L

Ideal Gas Law

• ideal gas - follows all assumptions of kinetic theory (particles have no volume, no attraction between particles)

• used to find one factor that effects gas when the other 3 factors are known– about the gas at one moment in time (no changing

conditions)

• only law that allows you to determine the amount of gas

• PV = nRT– R is ideal gas law constant

Ideal Gas Law

PV = nRT

What is the volume (in liters) occupied by 49.8 g of HCl at STP?

PV = nRT

V = nRTP

T = 0°C + 273 = 273 K

P = 1 atm

n = 49.8 g x 1 mol HCl36.45 g HCl

= 1.37 mol

V =1 atm

1.37 mol x 0.08206 x 273 K L•atmmol•K

V = 30.6 L

Remember: n = moles!

R = 0.08206 L•atm/mol•K

Ideal Gas Law

PV = nRT

What is the pressure occupied by 28.0 g of O2 at 15.0°C and 2.50 L?

PV = nRT

P = nRTV

T = 15°C + 273 = 288 K

V = 2.50 L

n = 28.0 g x 1 mol O2

32.0 g O2

= 0.875 mol

P =2.50 L

0.875 mol x 0.08206 x 288 K L•atmmol•K

P = 8.27 atm

Remember: n = moles!

R = 0.08206 L•atm/mol•K