Pressure Conversions 1 atm = 1.01325 x 105 Pa 1 bar = 1 x 105 Pa 1 millibar (mb) = 100 Pa 1 atm = 1.01325 bar 1 atm = 760 torr 1 torr = 1 mm Hg

Section 8.7, pp. 417-422

Boyle’s Law 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 P1V1 = P2V2

A gas occupying a volume of 725mL at a pressure of 0 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) = (V2)(0.541 atm) V2 = 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 V1 V2 n1 n2 =

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 P1 P2 T1 T2 =

An aerosol can is under a pressure of 3. 00 atm at 25C 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

Charles’ Law 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 V1 V2 T1 T2 =

A sample of neon gas has a volume of 752mL at 25C 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 n  = constant Such that P1V1 P2V2 T1 T2 PV T =

A He filled balloon has a volume of 50. 0L at 15C and 820mmHg 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

PV = nRT Ideal Gas Law! PV n  = constant T 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) 0oC 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 22oC) PV = nRT (1 atm) (0.6 L) = n (0.08206 L.atm/mol.K) (295 K) n = 0.025 mols