�ݺ�ߣ

Gas Laws Review
Created By: Blake Lubbers

Boyle’s Law
O The relationship between temperature and
pressure
O (P1)(V1)=(P2)(V2)
O When one goes down, the other must go
up
O Ex) P1) 1 torr, V1) 4 Liters, P2) 2 torr, V2)
? 1*4=2*?
O ?=2 Liters

Charles’s Law
(Yes I’ll be grammatically correct)

O Relationship between volume and
temperature
O Pressure and amount of gas is constant
O V1/T1=V2/T2
O One goes up, the other must go up as well
and vice versa
O Temperature Must be in Degrees Kelvin

Charles’s Law
(Continued…)

O Example
O 2L of a gas at 20◦C is put in to a 1300 mL
container. Find the new T in ◦C.
O First convert L to mL and ◦C to ◦K (2L =
2000 mL, 20◦C = 293◦K)
O 2000/293=1300/?
O T2= (1300)(293)/2000
O 190.45◦K = -82.55◦C

Gay-Lussac’s Law
( )
Ha ha…Gay-Lussac

O Relationship between temperature and
pressure
O Volume and amount remain constant
O P1/T1=P2/T2

Gay-Lussac’s Law (Continued)
O Example
O A 2L container at 30◦C has a pressure of 9
atm. What is the temperature at 4000
torr?
O 9 atm/1 * 760 torr/1 atm
O 6840/303 = 4000/T2
O T2=177.2◦K
O -95.81◦C
O -140.5◦F

Dalton’s Law
O Mixture of gasses
O Easiest to understand
O Simply adding together everything you
have

Dalton’s Law (Continued)
O Example
O A 4L container at 40◦C has O2 at 3 atm
and N2 at 4 atm. Find the total pressure.
O Total pressure = Partial O2 + Partial N2
O 3 atm + 4 atm = 7 atm

Dalton’s Law Backwards
O 6 particles of gas are pressing against a
solid object at 100 atm. What is the
individual atm for each gas if there are 4
particles of Gas A and 6 particles of Gas
B?
O Total of one gas over total of both gasses
O 6/10= 0.6 4/10= 0.4
O Total atm times partial atm
O Gas A- (100)(.6)=60 atm, Gas B-
(100)(.4)= 40 atm

Finding Partial Pressure
O Total pressure of a balloon is 3000 torr.
Find the partial pressure of each gas in
the balloon. (Gas A- 5 particles, Gas B- 3
particles, Gas C- 4 particles)
O 5/12=.417 (.417)(3000)=1251 torr
O 3/12=.25 (.25)(3000)=750 torr
O 4/12=.33 (.33)(3000)=990 torr

Graham’s Law
O Effusion of gasses (Diffusion through a
hole)
O Rate= A/B= √molecular weight B/
√molecular weight A

Graham’s Law (Continued)
O How much faster is H2(A) than CO2(B)?
O Rate H2/CO2 = √44/√2= 4.7 so H2 is 4.7
times faster than CO2
O Alternatively, CO2 is .213 times as fast as
H2.

Solving for and unknown gas
O An unknown gas(A) is 10X slower than
He(B), find the molecular weight.
O Rate= ?/He= √He/ √?
O 1/10= √4/ √X
O √X=10 √4
O (√X)= (20)^2
O X=400

Combined Gas Law
O Combination of Boyle’s, Charles’s, and
Gay-Lussac’s Laws
O (P1•V1)/ T1 = (P2•V2)/ T2.
P = pressure
V = volume
T = temperature

Ideal Gas Law
O PV=nRT
O P=Pressure in ATM
O V=Volume in Liters
O n=number of moles of the gas you're
dealing with
O R=.0821
O T= Temperature in degrees Kelvin
O All but one will be given to you in the
problem, you have to solve for the last
one

�ݺ�ߣ

Gas laws review

More Related Content

Gas laws review