ib chemistry on ideal gas equation, rmm determination of volatile liquid or gas
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Ideal Gas Equation
PV = nRT (n, T fix) PV = constant
V = constant/P
V ∝ 1/p
Charles’s Law
PV = nRT
4 different variables → P, V, n, T
Avogadro’s Law
T = Absolute Temperature in K
P1V1 = P2V2 V1 = V2
T1 T2
PV = nRT (n, V fix) P = constant x T
P ∝ T
V1 = V2
n1 n2
R = universal gas constant Unit - 8.314Jmol-1K-1 or 0.0821 atm L mol-1 K-1
n = number of moles
V = Volume gas Unit – dm3 or m3
P = Pressure Unit – Nm-2/Pa/kPa/atm
PV = nRT Fix 2 variables → change to different gas Laws
Boyle’s Law Pressure Law
P1 = P2
T1 T2
PV = nRT (P, T fix) V = constant x n
V ∝ n
PV = nRT (n ,P fix) V = constant x T V = constant
T V ∝ T
PV = nRT (n, T fix) PV = constant
V = constant/P
V ∝ 1/p
Charles’s Law Avogadro’s Law
PV = nRT (n ,P fix) V = constant x T V = constant
T V ∝ T
P1V1 = P2V2 V1 = V2
T1 T2
PV = nRT (n, V fix) P = constant x T
P ∝ T
V1 = V2
n1 n2
PV = nRT Fix 2 variables → change to different gas Laws
Pressure Law
P1 = P2
T1 T2
PV = nRT (P, T fix) V = constant x n
V ∝ n
Boyle’s Law PV = nRT (n, T fix) PV = constant V = constant P • V inversely proportional to P
V ∝ 1 P P1V1 = P2V2
Boyle’s Law Lab Simulator
Video on Boyle’s Law
Boyle’s Law
PV = nRT (n, T fix) PV = constant
V = constant/P
V ∝ 1/p
Charles’s Law Avogadro’s Law
PV = nRT (n ,P fix) V = constant x T V = constant
T V ∝ T
P1V1 = P2V2 V1 = V2
T1 T2
PV = nRT (n, V fix) P = constant x T
P ∝ T
V1 = V2
n1 n2
PV = nRT Fix 2 variables → change to different gas Laws
Pressure Law
P1 = P2
T1 T2
PV = nRT (P, T fix) V = constant x n
V ∝ n
Charles’s Law Lab Simulator
Video on Charles’s Law
Boyle’s Law
Charles’s Law PV = nRT (n, P fix) V = constant x T
• V directly proportional to T
V ∝ T V1 = V2
T1 T2
Temp increase ↑ → kinetic energy increase ↑ → collision bet particles with container increase ↑ → volume increase ↑
PV = nRT (n, T fix) PV = constant
V = constant/P
V ∝ 1/p
Charles’s Law Avogadro’s Law
PV = nRT (n ,P fix) V = constant x T V = constant
T V ∝ T
P1V1 = P2V2 V1 = V2
T1 T2
PV = nRT (n, V fix) P = constant x T
P ∝ T
V1 = V2
n1 n2
PV = nRT Fix 2 variables → change to different gas Laws
Pressure Law
P1 = P2
T1 T2
PV = nRT (P, T fix) V = constant x n
V ∝ n
Pressure Law Lab Simulator
Video on Pressure Law
Boyle’s Law
Pressure Law PV = nRT (n, V fix) P = constant x T
P directly proportional to T P ∝ T
P1 = P2
T1 T2
Temp increase ↑ → kinetic energy increase ↑ → collision bet particles with container increase ↑ → pressure increase ↑
PV = nRT (n, T fix) PV = constant
V = constant/P
V ∝ 1/p
Charles’s Law Avogadro’s Law
PV = nRT (n ,P fix) V = constant x T V = constant
T V ∝ T
P1V1 = P2V2 V1 = V2
T1 T2
PV = nRT (n, V fix) P = constant x T
P ∝ T
V1 = V2
n1 n2
PV = nRT Fix 2 variables → change to different gas Laws
Pressure Law
P1 = P2
T1 T2
PV = nRT (P, T fix) V = constant x n
V ∝ n
Avogadro Law Lab Simulator
Video on Avogadro Law
Boyle’s Law
Avogadro Law PV = nRT (P, T fix) V = constant x n
V directly proportional to n
V ∝ n V1 = V2
n1 n2
• 1 mole of any gas at fix STP (Std Temp/Pressure) • occupies a volume of 22.4dm3/22400cm3/24L
Avogadro’s Law
http://leifchemistry.blogspot.kr/2011/01/molar-volume-at-stp.html
Gas Helium Nitrogen Oxygen
Mole/mol 1 1 1
Mass/g 4.0 28.0 32.0
Pressure/atm 1 1 1
Temp/K 273 273 273
Vol/L 22.4L 22.4L 22.4L
Particles 6.02 x 1023 6.02 x 1023
6.02 x 1023
22.4L
“ equal vol of gases at same temperature/pressure contain equal numbers of molecules”
T – 0C (273.15K)
Unit conversion
1 atm = 760 mmHg/Torr = 101 325Pa(Nm-2) =101.325kPa 1m3 = 103 dm3 = 106cm3
1dm3 = 1 litre
P - 1 atm = 760 mmHg = 101 325Pa (Nm-2) = 101.325kPa
Standard Molar Volume Standard Temp/Pressure
“molar volume of all gases the same at given T and P” ↓
22.4L
22.4L 22.4L
Video on Avogadro’s Law
1 mole
gas
PV = nRT (n, T fix) PV = constant
V = constant/P
V ∝ 1/p
Charles’s Law Avogadro’s Law
PV = nRT (n ,P fix) V = constant x T V = constant
T V ∝ T
PV = nRT (n, V fix) P = constant x T
P ∝ T
PV = nRT Fix 2 variables → change different gas Laws
Pressure Law
PV = nRT (P, T fix) V = constant x n
V ∝ n
Boyle’s Law
Combined Gas Law
Boyle’s Law Charles’s Law
V ∝ 1 P
V ∝ T
Combined Boyle + Charles Law
PV = constant T PV = R T
Gas constant, R
V ∝ T P
P1V 1 = P2V2
T1 T2 3 different variables
Charles’s Law Boyle’s Law Pressure Law Avogadro’s Law
Combined Boyle Law + Charles Law Combined Gas Law
2 different variables
2 different variables 3 different variables
PV = nRT (n, T fix) PV = constant
V = constant/P
V ∝ 1/p
Charles’s Law Avogadro’s Law
PV = nRT (n ,P fix) V = constant x T V = constant
T V ∝ T
PV = nRT (n, V fix) P = constant x T
P ∝ T
PV = nRT Fix 2 variables → change different gas Laws
Pressure Law
PV = nRT (P, T fix) V = constant x n
V ∝ n
Boyle’s Law
Boyle’s Law Charles’s Law
V ∝ 1 P
V ∝ n Boyle + Charles + Avogadro Law
Proportionality constant Gas constant, R
V ∝ n T P
4 different variables
Charles’s Law Boyle’s Law Pressure Law Avogadro’s Law
Boyle + Charles + Avogadro Law Ideal Gas Equation
2 different variables
PV = nRT
Ideal Gas Equation
Avogadro’s Law
V ∝ T
PV = n R T
Charles’s Law Pressure Law
PV = nRT
Avogadro’s Law Boyle’s Law
V ∝ 1 P
V ∝ T P ∝ T V ∝ n
When n = 1 mol – Gas constant, R is 8.31 JK-1mol-1or NmK-1
For 1 mole – PV = RT For n mole – PV = nRT P1V 1 = P2V2
T1 T2
PV = nRT
Ideal Gas Equation Combined Gas Law
+
+
2 different variables
3 different variables
4 different variables
PV = nRT
R = P V n T
n 1 mol
Temp/T oC → 273K
Pressure/P 101 325 Pa(Nm-2)
Volume/V 22.4dm3 → 22.4 x 10-3 m3
R = 101325 x 22.4 x 10-3
1 x 273
R = 8.31 JK-1mol-1or NmK-1
Find R (Universal Gas Constant)
at molar volume
n = 1 mol
T = 273K
P = 101325Pa/Nm-2
V = 22.4 x 10-3m3
R = ?
Value of gas constant, R (Universal Gas Constant) at molar volume
Different Units Used
Volume/V 22.4dm3 → 22.4 x 10-3 m3
PV = nRT Pressure/P
101 325 Pa(Nm-2)
Temp/T oC → 273K
n 1 mol
R = P V n T
R = 101325 x 22.4 x 10-3
1 x 273 R = 8.31 JK-1mol-1or NmK-1
PV = nRT
R = P V n T
n 1 mol
Temp/T oC → 273K
Volume/V 22.4L
Pressure/P 1 atm
R = 1 x 22.4
1 x 273 R = 0.0821 atmLmol-1K-1
1 atm ↔ 760 mmHg/Torr ↔ 101 325Pa/Nm-2 ↔ 101.325kPa 1m3 ↔ 103 dm3 ↔ 106cm3
1dm3 ↔ 1000cm3 ↔ 1000ml ↔ 1 litre x 103 x 103
cm3 ↔ dm3 ↔ m3
x 10-3 x 10-3
Unit conversion
Different Units Used
Determination of RMM using Ideal Gas Equation
Volatile Liquid (Propanone)
Volatile Gas (Butane)
Syringe Method Direct Weighing Direct Weighing
PV = nRT
Converted to gas
Heated
PV = n x R x T PV = mass x R x T M M = m x R x T P V
RMM calculated if- m, T, P, V, ρ are known
Density, ρ = m V
PV = n x R x T PV = mass x R x T M M = m x R x T V P M = ρ x R x T P
RMM = M
or n = mass M
Ideal Gas Equation
n = mass M
Determination of RMM (LIQUID) using Ideal Gas Equation
Direct Weighing
PV = nRT
PV = n x R x T PV = mass x R x T M M = m x R x T PV = 0.52 x 8.314 x 373 101325 x 2.84 x 10-4
= 56.33
Data Collection
RMM calculated - m, T, P, V are known
1. Cover the top with aluminium foil.
Procedure
2. Make a hole on aluminium foil
3. Record mass flask + foil
4. Pour 2 ml volatile liquid into flask
5. Place flask in water, heated to boiling Temp and record pressure
6. Vapour fill flask when heated
7. Cool flask in ice bath – allow vapour to condense to liquid
8. Take mass of flask + foil + condensed liquid
Mass flask 115.15 g
Mass flask + foil +
condensed vapour
115.67 g
Mass condensed vapour 0.52 g
Atmospheric pressure 101325 Pa
Temperature of boiling
water
100 0C →373K
Volume of flask 284 cm3 → 2.84 x 10-4 m3
Data Processing
Vol gas = Vol water = Vol water in flask = Mass water Assume density water = 1g/ml
Click here for lab procedure
Video on RMM determination
Syringe Method
PV = nRT
PV = n x R x T PV = mass x R x T M M = m x R x T PV = 0.12 x 8.314 x 371 100792 x 7.2 x 10-5 = 51.1
Data Collection
RMM calculated - m, T, P, V are known
1. Set temp furnace to 98C.
Procedure
2. Draw 0.2ml liquid into a syringe Record mass syringe + liquid.
5. Inject liquid into syringe
6. Liquid will vaporise , Record vol of heated vapour + air
4. Record vol of heated air.
Data Processing
Mass syringe + liquid
before injection
15.39 g
Mass syringe + liquid
after injection
15.27 g
Mass of vapour 0.12 g
Atmospheric Pressure 100792Pa
Temp of vapour 371 K
Volume heated air 7 cm3
Volume heated air +
vapour
79 cm3
Volume of vapour 72 – 7 = 72 cm3
72cm3 →7.2 x 10-5 m3
Click here for lab procedure
Video on RMM determination
Determination of RMM (LIQUID) using Ideal Gas Equation
Direct Weighing
PV = nRT
PV = n x R x T PV = mass x R x T M M = m x R x T PV = 0.65 x 8.314 x 294.75 99.17 x 2.76 x 10-4
= 58.17
Data Collection
RMM calculated - m, T, P, V are known
1. Fill a flask with water and invert it .
Procedure
2. Record pressure + temp of water
3. Mass of butane + lighter (initial)
4. Release gas into flask
6. Measure the vol gas
7. Mass of butane + lighter (final)
Mass butane + lighter
(initial)
87.63 g
Mass butane + lighter
(final)
86.98 g
Mass butane 0.65 g
Pressure gas 743.9 mmHg
760 mmHg → 101325 Pa
743.9mmHg → 99.17Pa
Temperature water 21.70C → 294.75K
Volume of gas 276cm3 → 2.76 x 10-4 m3
Data Processing
Click here for lab procedure
Total Pressure (atm) = partial P(butane) + partial P(H2O) P butane = P(atm) – P(H2O) = (760 – 19.32) mmHg P butane = 743.911 mmHg → 99.17Pa
Dalton’s Law of Partial Pressures: Total pressure of mix of gases = sum of the partial pressures of all the individual gases
5. Adjust water level in flask until the same as atm pressure
Determination of RMM (GAS) using Ideal Gas Equation
Video on RMM determination
RMM butane RMM butane Collection gas
Density of gas at stp is 2.78gdm-3. Find RMM of gas Answer 1 dm3 gas at stp = 1.78g 22.4 dm3 at stp = 1.78 x 22.4g (Molar Volume) = 39.87g (Molar Mass) RMM gas = 39.87
Mass of 1 dm3 gas is 1.96g at stp. Find RMM of gas Answer Mass of 1dm3 at stp = 1.96g Mass of 22.4dm3 at stp = 22.4 x 1.96g = 43.9g RMM = 43.9
IB Questions on Ideal Gas
1 2
3 4 Density of gas is 2.6gdm-3 , T- 25C and P - 101kPa Calculate RMM of gas Answer PV = n R T PV = m x R x T M M = m x R x T V P M = ρ x RT P Density = 2.6 gdm-3 → 2.6 x 103 gm-3
P = 101kPa → 101 x 103 Pa M = ρ x RT P M = (2.6 x 103) x 8.31 x (298) 101 x 103
= 63.7
Density ρ = m (mass) V(volume)
Density gas is 1.25gdm-3 at T- 25C and P- 101kPa. Calculate RMM of gas Answer Using PV = n R T n = m M PV = m x RT M M = m x RT V P M = ρ x RT = 1.25 x 103 x 8.31 x 298 P 1.01 x 103 = 30.6gmol-1
m = density, ρ = 1.25gdm-3
V
Calculate RMM of gas Mass empty flask 25.385g Mass flask filled gas = 26.017 Mass flask filled water = 231.985g Temp = 32C Pressure = 101kPa Answer Mass gas = (26.017 – 25.385) = 0.632g Vol flask = (231.985 – 25.385) = 206.6 x 10-6 m3
P = 101kNm-2 → 101 x 103 Nm-2
M = m x RT PV = 0.632 x 8.314 x 305 101 x 103 x 206.6 x 10-6 = 76.8
IB Questions on Ideal Gas
Find molar mass gas by direct weighing, at T-23C and P- 97.7kPa Mass empty flask = 183.257g Mass flask + gas = 187.942g Mass flask + water = 987.560g Mass gas = (187.942 – 183.257) = 4.685g Vol gas = Vol water = Mass water = (987.560 – 183.257) = 804.303cm3
Vol gas = 804.303cm3 → 804.303 x 10-6m3
Pressure = 97.7kPa → 97700Pa Temp = 23C → (273 + 23) = 296K Assume density water = 1g/cm3
Answer PV = nRT PV = mass x R x T M M = mass x R x T PV = 4.685 x 8.314 x 296 97700 x 804.303 x 10-6
= 146.7g/ml
5 6
3.376g gas occupies 2.368dm3 at T- 17.6C, P - 96.73kPa. Find molar mass Answer PV = nRT PV = mass x RT M M = mass x R x T PV = 3.376 x 8.314 x 290.6 96730 x 2.368 x 10-3 = 35.61g/mol
Unit conversion
T – 17.6C → (273 = 17.6 = 290.6) P – 96.73kPa → 96730Pa
7
6.32 g gas occupy 2200cm3, T- 100C P -101kPa. Calculate RMM of gas
Answer PV = nRT n = PV RT n = (101 x 103) (2200 x 10-6) 8.31 x (100+273) n = 7.17 x 10-2 mol n = mass M RMM = mass n RMM = 6.32 7.17 x 10-2
= 88.15
8
Find empirical formula for composition by mass. S 23.7%, O 23.7%, CI 52.6% Density of its vapour at T- 70C and P- 98kNm-2 = 4.67g/dm3 What molecular formula?
Empirical formula - SO2CI2
PV = nRT PV = m x R x T M M = m x R x T V P Density ρ = m (mass) V (volume) M = ρ x RT P Density = 4.67gdm-3 → 4.67 x 103 gm-3
P = 98kN-2 → 9.8 x 104 Nm-2
M = (4.67 x 103) x 8.31 x (273 +70) 9.8 x 104
= 135.8 135.8 = n [ 32 + (2x16)+(2 x 35.5) ] 135.8 = n [ 135.8] n = 1 MF = SO2CI2
IB Questions on Ideal Gas
Element S O CI
Composition 23.7 23.7 52.6
Moles 23.7 32.1
= 0.738
23.7 16.0
= 1.48
52.6 35.5
= 1.48
Mole ratio 0.738 = 1 0.738
1
1.48 = 2 0.738
2
1.48 = 2 0.738
2
9
Answer Answer
A gas occupy at (constant P) • V - 125cm3
• T - 27C Calculate its volume at 35C
Answer: (Charles Law) V1 = V2 (constant P) T1 T2
125 = V2 (27+273) (35 + 273)) V2 = 128cm3
Find final vol, V2, gas at (constant T) compressed to P2 = 250kPa V1 - 100cm3
P1 - 100kPa V2 - ? P2 – 250kPa
Answer: (Boyle Law) p1V1 = p2V2 (constant T) 100 x 100 = 250 x V2
V2 = 40cm3
What volume (dm3) of 1 mol gas at P - 101325Nm-2 T - 25C
Answer: (Ideal gas eqn) pV = nRT V = nRT P V = 1 x 8.31 x (273 + 25) 101325 = 0.0244m3 = 24.4dm3
Find volume (m3) of 1 mol of gas at • T - 298K • P - 101 325Pa
Answer: (Ideal gas eqn) PV = nRT V = nRT P V = 1 x 8.314 x 298 101325 = 0.0244m3
Find volume (dm3) of 2.00g CO at • T → 20C • P → 6250Nm-2
Answer: (Ideal Gas Eqn) PV = nRT V = nRT P = 0.0714 x 8.314 x 293 6250 =0.0278m3 = 27.8dm3
IB Questions on Ideal Gas
T → (20 + 273) = 293K n → 2.00/28 = 0.0714 mol
Using PV = nRT (Ideal gas eqn) • Need to convert to SI units • 4 variables involved
3.0 dm3 of SO2 reacted with 2.0 dm3 of O2
2SO2(g) + O2(g) → 2SO3(g) Find volume of SO2 in dm3 at stp
Answer: (Avogadro Law) PV = nRT (at constant P,T) V ∝ n 2SO2(g) + 1 O2(g) → 2SO3(g) 2 mol 1 mol 2 mol 2 vol 1 vol 2 vol 3dm3 2dm3 ?
SO2 is limiting 2dm3 SO2 → 2dm3 SO3
3dm3 SO2 → 3dm3 SO3
Boyle, Charles, Avogadro Law
• no need to convert to SI units
• cancel off at both sides
• 2 variables involved
10 11 12
13 14 15
IB Questions on Ideal Gas
Combined gas Law
• no need to convert to SI units
• cancel off at both sides
• 3 variables involved
16 17 3
18 19
A syringe contains gas at V1 - 50cm3
P1 – 1atm T1 - 20C → 293K What volume , V2, if gas heated to V2 - ? T2 - 100C → 373K P2 - 5 atm
Answer: (Combine Gas Law) P1V1 = P2V2 T1 T2 1 x 50 = 5 x V2 293 373 V2 = 13cm3
Find volume fixed mass gas when its pressure and temp are double ?
Answer: (Combine Gas Law) Initial P1 → Final P2 = 2P1
Initial T1 → Final T2 = 2T1
Initial V1 → Final V2 = ? P1V1 = P2V2 T1 T2
P1V1 = 2P1V2 T1 2T1
V2 = V1 Volume no change
P and T double
Which change in conditions would increase the volume by x4 of a fix mass of gas?
Pressure /kPa Temperature /K
A. Doubled Doubled
B. Halved Halved
C. Doubled Halved
D. Halved Doubled
Answer: (Combine Gas Law) Initial P1 → Final P2 = 1/2P1
Initial T1 → Final T2 = 2T1
Initial V1 → Final V2 = ? P1V1 = P2V2 T1 T2
P1V1 = P1V2 T1 2 x 2T1
V2 = 4V1
Volume increase by x4
Fix mass ideal gas has a V1 = 800cm3 , P1, T1
Find vol, V2 when P and T doubled. V2 = ? P2 = 2P1
T2 = 2T1
Answer: (Combine Gas Law) Initial P1 → Final P2 = 2P1
Initial T1 → Final T2 = 2T1
Initial V1 800 → Final V2 = ? P1V1 = P2V2 T1 T2 P1 x 800 = 2P1V2 T1 2T1
V2 = 800
A. 200 cm3
B. 800 cm3
C. 1600 cm3
D. 3200 cm3
P halved T double
Ideal Gas Law Lab simulation
Ideal Gas Law Videos
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