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DO NOW: Complete on the BACK of the HW WS!

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Presentation on theme: "DO NOW: Complete on the BACK of the HW WS!"— Presentation transcript:

1 DO NOW: Complete on the BACK of the HW WS!
A gas has a pressure of 0.12 atm at 21.0 °C. What is the pressure at standard temperature? 2. If I have 5.6 L of gas in a piston at a pressure of atm and compress the gas until its volume is L, what will the new pressure inside the piston be? 3. If I have 45 liters of helium in a balloon at 250 C and increase the temperature of the balloon to C, what will the new volume of the balloon be?

2 Notes Unit: Gas Laws Combined Gas Law

3 After today you will be able to…
Explain the effect on gas properties using the Combined Gas Law Calculate an unknown pressure, temperature, or volume by solving algebraically

4 The Combined Gas Law The combined gas law is a single expression that combines Boyle’s, Charles’s, and Gay-Lussac’s Laws. This gas law describes the relationship between temperature, pressure, and volume of a gas. It allows you to do calculations where only the amount of gas is constant. RB + JC + JG-L = BFFs!

5 P1V1 P2V2 T1 T2 = The Combined Gas Law
Helpful hint: You are able to get which law you need by covering the variable that is not mentioned in the problem! There is no need to memorize 4 individual laws, just memorize the Combined Gas Law and you can derive all of the others! If there is no mention of pressure in the problem, cover P up and you are left with the relationship between T and V. (aka Charles’s Law!) If there is no mention of volume in the problem, cover V up and you are left with the relationship between T and P. (aka Gay-Lussac’s Law!) For example, if there is no mention of temperature in the problem, cover T up and you are left with the relationship between P and V. (aka Boyle’s Law!) P1V1 P2V2 T T2 =

6 The Combined Gas Law P1 V1 P2 V2 = T1 T2 P2 467mmHg = P1= V1= T1= P2=
A gas occupies 3.78L at 529mmHg and 17.2°C. At what pressure would the volume of the gas be 4.54L if the temperature is increased to 34.8°C? P1= V1= T1= P2= V2= T2= P1 V1 P2 V2 T T2 = 529mmHg 3.78L 17.2°C + 273= 290.2K (529mmHg) (3.78L) (P2) (4.54L) ? = (290.2K) (307.8K) 4.54L 467mmHg 34.8°C + 273= 307.8K P2 =

7 Questions? Complete WS

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