Pressure Conversions

• 1 atm = 1.01325 x 10 5 Pa • 1 bar = 1 x 10 5 Pa • 1 millibar (mb) = 100 Pa • 1 atm = 1.01325 bar • 1 atm = 760 torr • 1 torr = 1 mm Hg

Change volume (V) and the pressure (P) will change

(assuming that temperature and the number of molecules are constant)

Boyle’s Law PV = constant Pressure and Volume are

inversely

proportional P 1 V 1 = P 2 V 2

A gas occupying a volume of 725mL at a pressure of 0.970 atm is allowed to expand at constant temperature until its pressure reaches 0.541 atm. What is the final volume?

(725 ml)(0.970 atm) = (V 2 )(0.541 atm) V 2 = 1300 ml or 1.30 L

Change the amount of gas (n) and the volume (V) will change

(assuming that temperature and pressure constant)

Avogadro’s Law n  V Number of Moles and Volume are proportional V 1 = n 1 V 2 n 2

Change the temperature (T) and the pressure (P) will change

(assuming that the volume and number of moles are constant)

Gay-Lussac’s Law T  P Temperature and Pressure are proportional P 1 = T 1 P 2 T 2

An aerosol can is under a pressure of 3.00 atm at 25  C. Directions on the can caution the user to keep the can in a place where the temperature does not exceed 52  C. What would the pressure of the gas in the aerosol can be at 52  C?

3.00 atm = X atm 298 K 325 K X = 3.27 atm

Change the temperature (T) and the volume (V) will change

(assuming that the pressure and number of moles are constant)

Charles’ Law T  V Temperature and Pressure are proportional V 1 = T 1 V 2 T 2

A sample of neon gas has a volume of 752mL at 25  C.

What volume will the gas occupy at 50  C if the pressure remains constant?

752 ml X ml = 298 K 323 K X = 815 ml

What we find is that

everything is interrelated

… The combined Gas Law Such that n  PV T = constant P 1 V 1 T 1 = P 2 V 2 T 2

A He filled balloon has a volume of 50.0L at 15  C and 820mmHg. What volume will it occupy at 650mmHg and 10  C?

(50.0 L)(820 mmHg) = (X L)(650 mmHg) (288 K) (283 K) X = 62.0 L

n  PV T = constant

By defining the constant we can convert the proportionality into “workable” equation

PV = nRT

R is a constant which changes according to units, See Table 8.1 on page 401 R = 0.08206 L .

atm/mol .

K R = 8.314 J/mol .

K

Ideal Gas Conditions

• Negligible Interactions • Negligible Particle Size • High Temperature • Low Pressure

Standard Temperature and Pressure (STP)

0 o C 1 atm Under standard conditions, what is the volume of 1.00 mol of gas?

PV = nRT (1 atm)( V ) = (1 mol)(0.08206 L .

atm/mol .

K)(273 K) V = 22.4 L

How many moles of gas are in my 600 ml Pepsi bottle?

(assume that the room temperature is 22 o C) PV = nRT (1 atm)(0.6 L) = n (0.08206 L .

atm/mol .

K) (295 K) n = 0.025 mols