chem equations

Question 1

Boyle’s Law

Answer 1

• Gas law that state that P (atm) and V (L) are INVERSELY proportional under constant temp and mols

P1V1 = P2V2

Question 2

Charles’ Law

Answer 2

• Gas law that state that V (L) and temp (K) are DIRECTLY proportional under constant pressure and mols

V1/T1 = V2/T2

Question 3

Avogadro’s Law

Answer 3

• Gas law that state that V (L) and n (mols) are DIRECTLY proportional under constant temp and pressure

V1/n1 = V2/n2

Question 4

Gay-Lussac’s Law

Answer 4

• Gas law that state that P (atm) and V (L) are DIRECTLY proportional under constant volume and mols

P1/T1 = P2/T2

Question 5

Combined Gas Law

Answer 5

P1V1 / n1T1 = P2V2 / n2T2

Question 6

5 postulates of the Ideal Gas Law

Answer 6

1. Particles in a gas are in constant, random motion
2. Combined volume of particles in negligible
3. Particles exert no force on one another
4. Gas molecule collisions are completely elastic (no IMF)
5. All gases have the same average kinetic energy at a given temperature (avg kinetic energy of particles is proportional to the temp in Kelvins)

Question 7

Ideal Gas Law formula

Answer 7

PV = nRT

R can be either, depends on the units of pressure (atm or kPa)

Question 8

What is the Standard Temperature and Pressure (STP)?

Answer 8

Standard temp is 273 K, standard pressure is 1 atm, for an ideal gas

1 mol of gas at STP = 22.4 L

Question 9

Daltons Law of Partial Pressure

Answer 9

Ptotal = P1 + P2 + P3 +…
(sum of all the partial pressures of gases)

P1 = X1(Ptotal)
- X1 is the mole fraction of gas 1 (mols of gas 1/total mols)

Question 10

Gas density

Answer 10

d = m/V = PM/RT

Where:
d = density (p)
m = mass
V = voume
P = pressure
M = molar mass
R = ideal gas constant
T = temperature (K)

Question 11

Graham’s Law of Effusion

Answer 11

• says the rate of effusion of a gas is inversely proportional to the square root of its molar mass
• LIGHT gases diffuse and effuse much FASTER than heavier gases

r1/r2 = sqrt(M2/M1)

where:
r1 = effusion rate of gas 1
r2 = effusion rate of gas 2
M1 = molar mass of gas 1
M2 = molar mass of gas 2

Question 12

Freezing-Point Depression

Answer 12

• a drop in the minimum temp a which a substance freezes
• caused when a non-volatile solute is added
• means that a SOLUTION has a lower freezing point than a pure solvent

ΔTf = - kf i m

ΔTf = amount freezing point decreases
Kf = solvents freezing point depression constant
i = solute’s Van’t Hoff factor
m = molal concentration of solution

Question 13

Osmotic Pressure

Answer 13

the pressure it would take to stop osmosis

Π = iMRT

Π = osmotic pressure (atm)
i = van’t Hoff factor
M = molarity of solution
R = universal gas constant (0.0821 L atm/K mol)
T = temperature in Kelvin

Question 14

Boiling-point elevation

Answer 14

• the phenomenom that the boiling point of a liquid (a solvent) will be HIGHER when a non-volatile solute is added
• means that a solution has a higher bp than a pure solvent

ΔTb = kb i m

ΔTb = amount boiling point increases
kb = solvents boiling point elevation constant
i = solutes van’t hoff factor
m = molal concentration of solution

Question 15

Raoult’s Law

Answer 15

calculating new vapour pressure

PA = XA P°A
(subscript A)

PA = vapour pressure of solution
XA = mole fraction of solvent
P°A = pure vapour pressure of solvent

Question 16

Vapor pressure and boiling point relationship

Answer 16

Generally :
increase in vapor pressure = increase in bp

adding solute:
decrease vapour pressure = increase in bp

Question 17

molality

Answer 17

m = mols solute/kg of solvent

Question 18

molarity

Answer 18

M = mols solute/L of solution

solution is solute and solvent combined

Question 19

Beer’s Law

Answer 19

A = εlc

A = abosrbance
ε = molar absorptivity concept (M^-1 cm^-1)
l = path length
c = concentration of solution (M)