Kinetic Theory of Gases Maxwell Boltzmann Distribution Curve - Molecular speed/energies at constant temp - Molecule at low, most probable, root mean square speed - Higher temp –greater spread of energy to right (total area under curve the same) Straight line Curve line Vol gas – negligible IMF - (negligible) Low temp High temp Kinetic Theory simulation 2 2 1 mv KE Kinetic Theory of Gas 5 assumption - Continuous random motion, in straight lines - Perfect elastic collision - Ave kinetic energy directly proportional to abs temp ( E α T ) - Vol gas is negligible - Intermolecular forces attraction doesn’t exist
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IB Chemistry Real, Ideal Gas and Deviation from Ideal Gas behaviour
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Kinetic Theory of Gases
Maxwell Boltzmann Distribution Curve - Molecular speed/energies at constant temp
- Molecule at low, most probable, root mean square speed - Higher temp –greater spread of energy to right
(total area under curve the same)
Straight line Curve line
Vol gas – negligible IMF - (negligible)
Low temp
High temp
Kinetic Theory simulation
2
2
1mvKE
Kinetic Theory of Gas 5 assumption - Continuous random motion, in straight lines - Perfect elastic collision - Ave kinetic energy directly proportional to abs temp ( E α T ) - Vol gas is negligible - Intermolecular forces attraction doesn’t exist
Distribution of molecular speed, Xe, Ar, Ne, He at same temp At same temp • Xe, Ar, Ne and He have same Ave KE
• Mass He lowest – speed fastest • Mass Xe highest – speed slowest
He Ar Ne Xe
Why kinetic energy same for small and large particles? He – mass low ↓ - speed v high ↑ 2.2
1vmKE
Xe – mass high ↑ - speed v low ↓ 2.2
1vmKE
Kinetic energy SAME
Maxwell Boltzman Distribution Curve • Molecular speed/energy at constant Temp • Molecule at low, most probable and high speed • Higher temp –greater spread of energy to right • Area under curve proportional to number of molecules • Wide range of molecules with diff KE at particular temp • Y axis – fraction molecules having a given KE • X axis – kinetic energy/speed for molecule
2
2
1mvKE
Real Gas vs Ideal Gas
Deviation from ideal increase as
Molecule close together Forces of attraction exist
Kinetic energy low Molecule closer together – condensed Intermolecular forces stronger
Deviation of Real gas from Ideal
No forces attraction Forces attraction
1RT
PVPV constant at all
pressure, if temp constant
1RT
PV1
RT
PVor
Pressure increase ↑
Pressure exert on wall less ↓
1RT
PV
Temp decrease ↓
Temp ↓
PV = nRT Real Gas
Vol
Intermolecular forces attraction
Ideal Gas
Vol No intermolecular forces attraction
P
PV
Ideal gas
real gas
At Very High Press Vol gas is significant
At Low temp Molecules close together – condense Presence of intermolecular attraction Molecule will exert lower press at wall
1RT
PV
Deviation of Real gas from Ideal
At High Pressure Molecule close together Forces of attraction exist
Vol for molecule to move abt < observed vol because molecule occupy space. As pressure increase, free
space formolecule to move become smaller.
Vol gas significant at
high press
Vol used for calculation
Actual vol is less for gas to move
1RT
PV
Vol used in calculation is vol of container (too large)
Click here for notes from chemguide
At low temp- greater deviation
Negative deviation Presence of intermolecular attraction
Calculate RMM of gas Mass empty flask = 25.385 g Mass flask fill gas = 26.017 g
Mass flask fill water = 231.985 g Temp = 32C, P = 101 kPa
Find molar mass gas by direct weighing, T-23C , P- 97.7 kPa Mass empty flask = 183.257 g Mass flask + gas = 187.942 g Mass flask + water = 987.560 g Mass gas = (187.942 – 183.257) = 4.685 g Vol gas = Vol water = Mass water = (987.560 – 183.257) = 804.303 cm3
RMM determination
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 = 146.7
Vol gas = 804.303 cm3 = 804.303 x 10-6 m3
P = 97.7 kPa = 97700 Pa
Density water = 1g/cm3
M = m x RT PV = 0.632 x 8.314 x 305 101 x 103 x 206 x 10-6 = 76.8
m gas = (26.017 – 25.385) = 0.632 g
vol gas = (231.985 – 25.385) = 206 x 10-6 m3
X contain C, H and O. 0.06234 g of X combusted, 0.1755 g of CO2 and 0.07187 g of H2O produced.
Find EF of X
Element C H O
Step 1 Mass/g 0.0479 0.00805 0.006384
RAM/RMM 12 1 16
Step 2 Number moles/mol
0.0479/1 2 = 0.00393
0.00805/1 = 0.00797
0.006384/16 = 0.000393
Step 3 Simplest ratio 0.00393 0.000393
= 10
0.00797 0.000393
= 20
0.000393 0.000393
= 1
Conservation of mass Mass C atom before = Mass C atom after Mass H atom before = Mass C atom after
CHO + O2 CO2 + H2O
Mol C atom in CO2
= 0.1755 = 0.00393 mol 44 Mass C = mol x RAM C = 0.00393 x 12 = 0.0479 g
Mol H atom in H2O = 0.07187 = 0.0039 x 2 = 0.00797 mol 18 Mass H = mol x RAM H = 0.00797 x 1.01 = 0.00805 g
Mass of O = (Mass CHO – Mass C – Mass H) = 0.06234 – 0.0479 - 0.00805 = 0.006384 g
0.06234 g 0.1755 g 0.07187 g
Empirical formula – C10H20O1
Find EF for X with composition by mass. S 23.7 %, O 23.7 %, CI 52.6 % Given, T- 70 C, P- 98 kNm-2 density - 4.67g/dm3
What molecular formula?
Empirical formula - SO2CI2
Density ρ = m (mass) V (vol)
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 0.738
1
1.48 0.738
2
1.48 0.738
2 P
RTM
P
RT
V
mM
RTM
mPV
nRTPV
Density = 4.67 gdm-3 = 4.67 x 10-3 gm-3
M = (4.67 x 10-3) x 8.31 x (273 +70) 9.8 x 104
M = 135.8
135.8 = n [ 32 + (2 x 16)+(2 x 35.5) ] 135.8 = n [ 135.8] n = 1 MF = SO2CI2
P = 98 kN-2 = 9.8 x 104 Nm-2
3.376 g gas occupies 2.368 dm3 at T- 17.6C, P - 96.73 kPa. Find molar mass
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 = 35.61
Vol = 2.368 dm3 = 2.368 x 10-3 m3
P – 96.73 kPa → 96730Pa
T – 290.6K
6.32 g gas occupy 2200 cm3, T- 100C , P -101 kPa. Calculate RMM of gas
PV = nRT n = PV RT n = (101 x 103) (2200 x 10-6) 8.31 x ( 373 ) n = 7.17 x 10-2 mol
Vol = 2200 cm3 = 2200 x 10-6 m3
RMM = mass n RMM = 6.32 7.17 x 10-2
= 88.15
Sodium azide, undergoes decomposition rxn to produce N2 used in air bag
2NaN3(s) → 2Na(s) + 3N2(g)
Temp, mass and pressure was collected in table below i. State number of sig figures for Temp, Mass, and Pressure i. Temp – 4 sig fig Mass – 3 sig fig Pressure – 3 sig fig
Temp/C Mass NaN3/kg Pressure/atm
25.00 0.0650 1.08
ii. Find amt, mol of NaN3 present ii. iii. Find vol of N2, dm3 produced in these condition
RMM NaN3 – 65.02
molMol
RMM
massMol
00.102.60
0.65
P
nRTV
nRTPV
n = 1.50 mol
P – 1.08 x 101000 Pa = 109080 Pa
2NaN3(s) → 2Na(s) + 3N2(g)
T – 25.00 + 273.15 = 298.15K
2 mol – 3 mol N2
1 mol – 1.5 mol N2
33 1.340341.0
109080
15.29831.850.1
dmmV
V
P
nRTV
Density gas is 2.6 gdm-3 , T- 25C , P – 101 kPa Find RMM of gas
P
RTM
P
RT
V
mM
RTM
mPV
nRTPV
Density ρ = m (mass) V (vol)
M = (2.6 x 103) x 8.31 x (298) 101 x 103
M = 63.7
Sodium azide, undergoes decomposition rxn to produce N2 used in air bag
2NaN3(s) → 2Na(s) + 3N2(g)
Temp, mass and pressure was collected in table below
Temp/C Volume N2/L Pressure/atm
26.0 36 1.15
Find mass of NaN3 needed to produce 36L of N2
RMM NaN3 – 65.02
RT
PVn
nRTPV
1.1 x 65.02 = 72 g NaN3
P – 1.15 x 101000 Pa = 116150 Pa
2NaN3(s) → 2Na(s) + 3N2(g)
T – 26.0 + 273.15 = 299.15K
3 mol N2 – 2 mol NaN3
1.7 mol N2 – 1.1 mol NaN3
moln
n
7.1
15.29931.8
1036116150 3
Vol = 36 dm3 = 36 x 10-3 m3
Convert mole NaN3 → Mass /g
Density gas is 1.25g dm-3 at T- 25C ,P- 101 kPa. Find RMM of gas
P
RTM
P
RT
V
mM
RTM
mPV
nRTPV
Density ρ = m (mass) V (vol)
M = (1.25 x 103) x 8.31 x (298) 101 x 103
M = 30.6
PV
mRTM
RTM
mPV
nRTPV
Copper carbonate, CuCO3, undergo decomposition to produce a gas. Determine molar mass for gas X
CuCO3(s) → CuO(s) + X (g)
Temp, mass, vol and pressure was collected in table below
Temp/K Vol gas/ cm3 Pressure/kPa Mass gas/g
293 38.1 101.3 0.088
Find Molar mass for gas X
P – 101300 Pa
T – 293 K
Vol = 38.1 cm3 = 38.1 x 10-6 m3
5.55
101.38101300
29331.8088.06
M
M
Potassium chlorate, KCIO3, undergo decomposition to produce a O2. Find amt O2 collected and mass of KCIO3 decomposed