Boyle's or Charles'
Gay-Lussac or Dalton's
Ideal Gas Law or Combined Gas Law
Pressure and Temperature Conversions
Graham's Law and Mixed Bag
100
This law states that there is an inverse relationship between pressure and volume of a gas. P1V1 = P2V2
What is Boyle's Law?
100
Total Pressure = Pa + Pb + Pc + .... is the equation for
What is Dalton's Law?
100
This gas law uses P1V1/T1 = P2V2/T2
What is the Combined Gas Law?
100
What temperature is 25oC in Kelvin?
298 K 25 + 273 = 298
100
Compare the rates of effusion of carbon dioxide (CO2) gas and hydrochloric acid (HCl) gas.
(rate CO_2)/(rate HCl) = √(M_HCl )/√(M_CO2 ) = √36.5/√44 = 0.9 (or rate HCl/rate CO2 = 1.10)
200
This law states that there is a direct relationship between a gas's volume and its temperature in Kelvin V1/T1 = V2/T2
What is Charles' Law
200
P1/T1 = P2/T2 is the equation for what law?
What is Gay-Lussac's Law?
200
This Gas Law uses PV = nRt
What is the Ideal Gas Law?
200
Express 0.830 atm in kilopascals (kPa)
0.830 atm x 101.3 kPa/1 atm = 84.1 kPa
200
The equation for calculating pressure is
P = F/A
300
Use Boyle's Law to solve for the missing value: P1 = 350 torr V1 = 200 mL P2 = 700 torr V2 = ?
P1V1 = P2V2 (350 torr)(200 mL) = (700 torr)(V2) V2 = (350 torr)(200 mL)/(700 torr) = 100 mL
300
Three components of air are carbon dioxide (CO2), nitrogen (N2) and oxygen (O2). In a sample containing these gases 1 atm (or 760 torr), what is the partial pressure of oxygen if the pressure of CO2 is 0.285 torr and the pressure of N2 is 593.525 torr?
760 torr = 0.285 torr + 593.525 torr + P_O2 P_O2 = 760 - (0.285 + 593.525) torr P_O2 = 166.190 torr
300
The volume of a gas is 27.5 mL at 22.0 oC and 0.974 atm. What will the volume be at 15.0 oC and 0.993 atm?
P1V1/T1 = P2V2/T2 V2 = P1V1T2/P2T1 V2 = (0.974 atm)(27.5 mL)(288 K)/(0.933 atm)(295 K) V2 = 26.3 mL
300
Express 0.83 atm in millimeters of mercury (mm Hg)
0.83 atm x 760 mm Hg/1 atm = 631 mm Hg
300
Which law is represented by the expression V = kn?
What is Avogadro's Law?
400
A flask containing a volume of 155 cm^3 of hydrogen gas was collected under pressure of 22.5 kPa. What pressure would have been required for the volume of the gas to have been 90.0 cm^3?
P1V1 = P2V2 (22.5 kPa)(155 cm^3) = (P2)(90.0 cm^3) P2 = (22.5)(155)/90 = 38.8 kPa
400
A sample of hydrogen gas at 47 oC exerts a pressure of 0.329 atm. The gas is heated to 77 oC at constant volume. What will the new pressure be?
P1/T1 = P2/T2 P2 = P1T2/T1 = (0.329 atm)(350 K)/(320 K) = 0.360 atm
400
What pressure in atmospheres (atm), is exerted by 0.325 mol of hydrogen gas in a 4.08 L container at 35 oC?
PV = nRT P = nRT/V = (0.325 mol)(0.0821)(308 K)/4.08 L P = 2.01 atm
400
Convert -30oC to Kelving
-30 + 273 = 243 K
400
What is STP?
Standard Temperature (0oC) and Standard Pressure (1 atm)
500
A sample of air has a volume of 140.0 mL at 67oC. At what temperature would its volume be at 50.0 mL (constant pressure)?
V1/T1 = V2/T2 T1 = 67oC + 273 = 340K 140 mL/340 K = 50 mL/T2 T2 = (50)(340)/140 = 121 K
500
A sample of gas collected over water at a temperature of 35oC has a barometric pressure reading of 742.00 torr. What is the partial pressure of the dry gas?
P(total) = P(dry gas) + P(water) 742.00 = P(dry gas) + 42.2 (from table using 35oC) P(dry gas) = 742 - 42.2 torr = 699.8 torr
500
Calculate the volume, in liters occupied by 2.00 mol hydrogen gas (H2) at 300.0 K and 1.25 atm.
V = nRT/P = (2.00 mol)(0.0821)(300.0 K)/1.25 atm V = 39.4 L H2
500
A sample of hydrogen gas (H2) effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.
rate H2/rate unknown = SQR M (unknown)/SQR M (H2) rate H2/rate X = 9 9 = Molar Mass Unknown/Molar Mass H2 9 = Molar Mass Unknown/2.02 Molar Mass Unknown = 18.18 g/mol
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