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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
Kinetic Theory of Gases
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
Pressure Law
Ideal Gas Equation
PV = nRT PV = constant
V = constant/P
V ∝ 1/p
Charles’s Law
PV = nRT
4 diff variables → P, V, n, T
Avogadro’s Law
PV = nRT V = constant x T V = constant
T
V ∝ T
P1V1 = P2V2 V1 = V2
T1 T2
V1 = V2
n1 n2
R = gas constant Unit - 8.314 Jmol-1K-1
V = Vol gas Unit – m3
PV = nRT Fix 2 variables
↓ change to diff gas Laws
Boyle’s Law
n, T fix n, P fix n, V fix
PV = nRT V = constant x n
V ∝ n
P, T fix
P = Pressure Unit – Nm-2/Pa/kPa
n = number of moles
T = Abs Temp in K
Vol Pressure
Temp Vol Temp
Pressure Vol
n
PV = nRT P = constant x T
P ∝ T
P1 = P2
T1 T2
Ideal Gas Equation
PV = nRT (n, T fix) PV = constant
V = constant/P
V ∝ 1/p
PV = nRT
4 diff variables → P, V, n, T
P1V1 = P2V2
R = gas constant Unit - 8.314 Jmol-1K-1
V = Vol gas Unit – m3
PV = nRT
Boyle’s Law
n, T fix
P = Pressure Unit – Nm-2/Pa/kPa
n = number of moles
T = Abs Temp in K
Pressure
Vol
Boyle’s Law Lab Simulator
Video on Boyle’s Law
Ideal Gas Equation
Charles’s Law
PV = nRT
4 diff variables → P, V, n, T
PV = nRT V = constant x T V = constant
T
V ∝ T
V1 = V2
T1 T2
R = gas constant Unit - 8.314 Jmol-1K-1
V = Vol gas Unit – m3
PV = nRT
n, P fix
P = Pressure Unit – Nm-2/Pa/kPa
n = number of moles
T = Abs Temp in K
Vol
Temp
Charles’s Law Lab Simulator
Video on Charles’s Law
Ideal Gas Equation
PV = nRT
4 diff variables → P, V, n, T R = gas constant
Unit - 8.314 Jmol-1K-1
V = Vol gas Unit – m3
PV = nRT
n, V fix
P = Pressure Unit – Nm-2/Pa/kPa
n = number of moles
T = Abs Temp in K
Pressure
Temp
Pressure Law
PV = nRT (n, V fix) P = constant x T
P ∝ T
P1 = P2
T1 T2
Pressure Law Lab Simulator
Video on Pressure Law
Ideal Gas Equation
PV = nRT
4 diff variables → P, V, n, T R = gas constant
Unit - 8.314 Jmol-1K-1
V = Vol gas Unit – m3
PV = nRT
P, V fix
P = Pressure Unit – Nm-2/Pa/kPa
n = number of moles
T = Abs Temp in K
Vol
n
Video on Pressure Law
Avogadro’s Law
PV = nRT V = constant x n
V ∝ n
V1 = V2
n1 n2
Avogadro Law Lab Simulator
Video on Avogadro Law
Avogadro’s Law
Gas Helium Nitrogen Oxygen
Mole/mol 1 1 1
Mass/g 4.0 28.0 32.0
Press /atm 1 1 1
Temp/K 273 273 273
Vol/L 22.7L 22.7L 22.7L
Particles 6.02 x 1023 6.02 x 1023
6.02 x 1023
22.7L
“ equal vol of gases at same temp/press contain equal numbers of molecules”
T – 0C (273.15 K)
Unit conversion
1 m3 = 103 dm3 = 106 cm3
1 dm3 = 1 litre
Standard Molar Volume
“molar vol of all gases same at given T and P” ↓ 22.7L
22.7L
Video on Avogadro’s Law
1 mole
gas
• 1 mole of any gas at STP (Std Temp/Press) • occupy a vol of 22.7 dm3/22 700 cm3
P - 1 atm = 760 mmHg = 100 000 Pa (Nm-2) = 100 kPa
22.7L
Unit conversion
1 atm ↔ 760 mmHg ↔ 100 000 Pa ↔ 100 kPa 1m3 ↔ 103 dm3 ↔ 106 cm3
1 dm3 ↔ 1000 cm3 ↔ 1000 ml ↔ 1 litre x 103 x 103
cm3 dm3 m3
x 10-3 x 10-3
Pressure Law
Ideal Gas Equation
PV = nRT PV = constant
V = constant/P
V ∝ 1/p
Charles’s Law Avogadro’s Law
PV = nRT V = constant x T V = constant
T
V ∝ T
P1V1 = P2V2 V1 = V2
T1 T2
V1 = V2
n1 n2
PV = nRT
Boyle’s Law
n, T fix n, P fix n, V fix
PV = nRT V = constant x n
V ∝ n
P, T fix
PV = nRT P = constant x T
P ∝ T
P1 = P2
T1 T2
Combined Boyle + Charles + Avogadro
2
22
1
11
T
VP
T
VP
Combined Boyle + Charles
nRTPV
P
nTV
Find R at molar vol n = 1 mol T = 273K
P = 100 000 Pa
V = 22.7 x 10-3 m3
R = ? R = 8.31 JK-1 mol-1
nT
PVR
T = 273K
V = 22.7 x 10-3 m3
P = 100 000 Pa
Volatile Liquid (Propanone)
Volatile Gas (Butane)
Syringe Method Direct Weighing
Direct Weighing
Heated – convert to gas
RMM calculated - m, T, P, V, ρ are known
n = mass M
P
RTM
P
RT
V
mM
RTM
mPV
nRTPV
Density ρ = m (mass) V (vol) PV
mRTM
RTM
mPV
nRTPV
Molar mass
RMM using Ideal Gas Eqn
PV = nRT
Direct Weighing
PV = nRT 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
1. Cover top with aluminium foil.
2. Make a hole on aluminium foil
3. Record mass flask + foil
4. Pour 2 ml volatile liq to flask
5. Place flask in water, heat to boiling Temp and record press
6. Vapour fill flask when heat
7. Cool flask in ice bath –allow vapour to condense to liquid
8. Take mass flask + foil + liquid
Mass flask + foil 115.15 g Mass flask + foil + condensed vapour
115.67 g
Mass condensed vapour
0.52 g
Pressure
101325 Pa
Temp of boiling water
100 0C
373K Vol of flask 284 cm3
2.84 x 10-4 m3
Data Processing
Vol gas =Vol water in flask = Mass water Assume density water = 1 g/ml
Click here for lab procedure
Video on RMM determination
RMM (LIQUID) using Ideal Gas Eqn
Procedure
Data Collection
Direct Weighing
1. Fill flask with water and invert it .
2. Record press + temp of water
3. Mass of butane + lighter (ini)
4. Release gas into flask
6. Measure vol gas
7. Mass of butane + lighter (final)
Total Press (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 Press: Total press of mix of gas = sum of partial press of all individual gas
5. Adjust water level in flask until the same as atm pressure
RMM butane RMM butane Collection gas
RMM (GAS) using Ideal Gas Eqn
Procedure Data Collection
Mass butane + lighter 87.63 g Mass butane + lighter
(final) 86.98 g
Mass butane
0.65 g
Pressure
99.17 Pa
Temp of boiling water
21.7 0C
294 K Vol of flask 276 cm3
2.76 x 10-4 m3
Data Processing
PV = nRT PV = mass x R x T M M = m x R x T PV = 0.65 x 8.314 x 294 99.17 x 2.76 x 10-4
= 58.17
Syringe Method
1. Set temp furnace to 98C.
2.Put 0.2ml liq into a syringe 3. Record mass syringe + liq
5. Inject liq into syringe
6. Liq will vaporise , Record vol of heat vapour + air
4. Record vol of heated air.
Mass syringe + liq bef injection
15.39 g
Mass syringe + liq after injection
15.27 g
Mass of vapour 0.12 g
Pressure 100792Pa
Temp of vapour 371 K
Vol heated air 7 cm3
Vol heated air + vapour 79 cm3
Vol of vapour 72 – 7 = 72 cm3
7.2 x 10-5 m3
Video on RMM determination
RMM (LIQUID) using Ideal Gas Eqn
Data Collection Procedure
Data Processing
PV = nRT 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
P = 101 kNm-2 = 101 x 103 Nm-2
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
KCIO3
Temp/K Vol gas/ dm3 Pressure/kPa
299 0.250 101.3 2KCIO3(s) → 2KCI(s) + 3O2 (g)
RT
PVn
nRTPV
2
3
.010.0
29931.8
10250.0101300
Omoln
n
Vol = 0.250 dm3 = 0.250 x 10-3 m3
P – 101300 Pa
Convert mole KCIO3 → Mass
2KCIO3 → 2KCI + 3O2
2 mol – 3 mol O2
0.0066 mol – 0.01 mol O2
0.0066 x 122.6 = 0.81 g KCIO3
RMM KCIO3 – 122.6
Gas occupy at (constant P) V – 125 cm3 , T - 27 C
Find its vol at 35 C
V1 = V2 (constant P) T1 T2
↓ 125 = V2
(27+273) (35 + 273) ↓
V2 = 128 cm3
Find final vol, V2, at (constant T) compressed to P2 = 250 kPa V1 - 100 cm3 , P1 – 100 kPa V2 - ? P2 – 250 kPa
p1V1 = p2V2 (constant T) ↓
100 x 100 = 250 x V2
↓ V2 = 40 cm3
What vol (dm3) of 1 mol gas at P - 101325 Pa, T - 25C
pV = nRT V = nRT P V = 1 x 8.31 x (273 + 25) 101325 = 0.0244m3 = 24.4dm3
Find vol (m3) of 1 mol of gas at T - 298K, P - 101 325Pa
PV = nRT V = nRT P V = 1 x 8.314 x 298 101325 = 0.0244 m3
Find vol (dm3) of 2.00g CO at T - 20C, P - 6250Nm-2
PV = nRT V = nRT P = 0.0714 x 8.314 x 293 6250 = 0.0278 m3 = 27.8dm3
IB Questions on Ideal Gas
T → 293K
3.0 dm3 of SO2 react with 2.0 dm3 of O2
2SO2(g) + O2(g) → 2SO3(g) Find vol of SO2 , dm3 at stp
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
n → 2.00/28 = 0.0714 mol
A syringe contains gas at V1 – 50 cm3 , P1 – 1 atm, T1 - 293K
What vol , V2, if gas heat to V2 ? T2 - 373 K, P2 - 5 atm
Find vol of gas when its press and temp are double ?
Volume no change
Which change in conditions would increase vol by x4 of a fix mass of gas?
Pressure /kPa Temperature /K
A. Doubled Doubled
B. Halved Halved
C. Doubled Halved
D. Halved Doubled
Initial P1 → Final P2 = 1/2P1
Initial T1 → Final T2 = 2T1
Initial V1 → Final V2 = ? Vol increase by x4
Fix mass ideal gas has a V1 = 800cm3 , P1, T1
Find vol, V2 when P and T doubled. P2 = 2P1
T2 = 2T1
Initial P1 → Final P2 = 2P1
Initial T1 → Final T2 = 2T1
Initial V1 800 → Final V2 = ?
A. 200 cm3
B. 800 cm3
C. 1600 cm3
D. 3200 cm3
3
2
2
2
22
1
11
13
373
5
293
501
cmV
V
T
VP
T
VP
12
1
21
1
11
2
22
1
11
2
2
VV
T
VP
T
VP
T
VP
T
VP
12
1
21
1
11
2
22
1
11
4
22
VV
T
VP
T
VP
T
VP
T
VP
3
2
1
21
1
1
2
22
1
11
800
2
2800
cmV
T
VP
T
P
T
VP
T
VP
Fix mass ideal gas has a V1 = 1dm3, P1, T1
Find V2 ,when T doubled (x2), P tripled (x3) V2 = ?, P2 = 3P1, T2 = 2T1
Initial P1 → Final P2 = 3P1
Initial T1 → Final T2 = 2T1
Initial V1 = 1 dm3 → Final V2 =?
Fix mass ideal gas has a V1 = 2 dm3 , P1, T1
Find V2 ,when T double (x2), P quadruple ( x4 ) V2 = ?, P2 = 4P1, T2 = 2T1
Initial P1 → Final P2 = 4P1
Initial T1 → Final T2 = 2T1
Initial V1 2dm3 → Final V2 = ?
Fix mass ideal gas has a P1 = 40 kPa , V1, T1
Find P2 of gas when V and T doubled. P2 = ?, V2 = 2V1, T2 = 2T1
Initial V1 → Final V2 = 2V1
Initial T1 → Final T2 = 2T1
Initial P1 =40 → Final P2 = ?
3/2
2
31
2
1
21
1
1
2
22
1
11
V
T
VP
T
P
T
VP
T
VP
3
2
1
21
1
1
2
22
1
11
1
2
42
dmV
T
VP
T
P
T
VP
T
VP
40
2
240
2
1
12
1
1
2
22
1
11
P
T
VP
T
V
T
VP
T
VP
What conditions would one mole CH4, occupy the smallest vol?
A. 273 K and 1.01×105 Pa B. 273 K and 2.02×105 Pa C. 546 K and 1.01×105 Pa D. 546 K and 2.02×105 Pa
PV = nRT V = nRT P = low T, high P
Find total vol and composition of remaining gas 10cm3 ethyne react with 50cm3 hydrogen
C2H2(g) + 2H2 (g) → C2H6 (g) at stp
PV = nRT (at constant P,T) V ∝ n C2H2 (g) + 2H2(g) → C2H6 (g)
1 mol 2 mol 1 mol 1 vol 2 vol 1 vol 10cm3 20cm3 10cm3
C2H6 = 10cm3 produced H2 = 50-20 =30 cm3 remain (excess)
Which conditions does a fix mass of an ideal gas have greatest vol?
Temperature Pressure
A. low low
B. low high
C. high high
D. high low
PV = nRT V = nRT P = high T, low P
Find vol of 6 g of chlorine at 27oC and 101 kPa.
pV = nRT T = 300K, n = 6/71 = 0.08451 mol CI2, Mr(Cl2)= 71 P = 101000 Pa.
3002087.0
101000
30031.808451.0
mV
V
P
nRTV
5 litre container contain 0.5 kg butane gas (C4H10). Cal press at 25oC.
Mr(C4H10)= 58, 0.5kg = 500g moles n = 500/58 = 8.621, T = 298K 5 litre = 5 dm3 = 5 x 10-3 m3
kPaP
P
V
nRTP
4272
105
29831.8621.83
P = 202600 Pa n = 0.050 mol T = 400K V = ? m3
What vol need to store 0.050 moles of helium, P= 202.6 kPa , T= 400 K
V = 7.5 x 10-3 m3 molar mass (H2) =2.016 g mol-1 n = 20.16 ÷ 2.016 = 10 mol T = 293 K
What press exerted by 20.16 g hydrogen in 7.5 L cylinder at 20oC?
300082.0
202600
40031.8050.0
mV
V
P
nRTV
nRTPV
kPaP
P
V
nRTP
nRTPV
3248
105.7
29331.8103
50 L fill with argon to a press of 101 kPa at 30oC. How many moles of argon ?
P = 101000Pa V = 50 x 10-3 m3 n = ? mol T = 303 K
moln
n
RT
PVn
nRTPV
2
30331.8
1050101000 3
What temp does a 250 mL cylinder containing 0.40 g helium need to be cool to a press of 253.25 kPa?
P = 253250 Pa V = 0.250 x 10-3 m3 mass = 0.40 g , molar mass (He) = 4.003 n = 0.40 ÷ 4.003 = 0.1 mol
KT
T
nR
PVT
nRTPV
15.76
31.81.0
10250.0253250 3