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
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
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, 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
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
1 2 3
4 5 6
IB Questions on Ideal Gas
Combined gas Law
• no need to convert to SI units
• cancel off at both sides
• 3 variables involved
7 8 3
9 10
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
A. 1 dm3
B. 2 dm3
C. 3 dm3
D. 4 dm3
IB Questions on Ideal Gas
Fix mass ideal gas has a V1 = 1dm3
P1
T1
Find V2 ,when T doubled (x2), P tripled (x3) V2 = ? P2 = 3P1
T2 = 2T1
Answer: (Combine Gas Law) Initial P1 → Final P2 = 3P1
Initial T1 → Final T2 = 2T1
Initial V1 = 1dm3 → Final V2 =? P1V1 = P2V2 T1 T2 P1 x 1 = 3P1 x V2 T1 2T1
V2 = 2/3
A. 1/3
B. 2/3
C. 3/2
D. 1/6
11 Fix mass ideal gas has a V1 = 2dm3 P1
T1
Find V2 ,when T double (x2), P quadruple.(x4) V2 = ? P2 = 4P1
T2 = 2T1
Answer: (Combine Gas Law) Initial P1 → Final P2 = 4P1
Initial T1 → Final T2 = 2T1
Initial V1 2dm3 → Final V2 = ? P1V1 = P2V2 T1 T2 P1 x 2 = 4P1 x V2 T1 2T1
V2 = 1 dm3
12
Fix mass ideal gas has a P1 = 40kPa V1
T1
Find P2 of gas when V and T doubled. P2 = ? V2 = 2V1
T2 = 2T1
Answer: (Combine Gas Law) Initial V1 → Final V2 = 2V1
Initial T1 → Final T2 = 2T1
Initial P1 =40 → Final P2 = ? P1V1 = P2V2 T1 T2 40 x V1 = P2 x 2V1 T1 2T1
P2 = 40
A. 10kPa
B. 20kPa
C. 40kPa
D. 480kPa
13
Combined gas Law
• no need to convert to SI units
• cancel off at both sides
• 3 variables involved
IB Questions on Ideal Gas
Calculate total volume and composition of remaining gas 10cm3 ethyne react with 50cm3 hydrogen to produce ethane C2H2(g) + 2H2(g) → C2H6(g) at stp
Answer: (Avogadro Law) 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 =30cm3 remains (excess)
Which conditions does a fix mass of an ideal gas have greatest volume?
Temperature Pressure
A. low low
B. low high
C. high high
D. high low
Answer: (Ideal Gas Eqn) PV = nRT V = nRT P = high T, low P
What conditions would one mole of, CH4, occupy the smallest volume?
Answer: (Ideal Gas Eqn) PV = nRT V = nRT P = low T, high P
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
14 15
16
Calculate the mass for a) 2/3 mole of aluminium atoms b) 0.08 mole of C6H8O6 molecules c) 0.125 mole Mg(OH)2
Answer: a) 1 mole Al atoms → 27g
2/3 mole AI atoms → 2/3 x 27 g → 18g
b) 1 mole, C6H8O6 → 6(12) + 8(1) + 6(16) = 176g 0.08 mole C6H8O6 → 0.08 x 176 g = 14.08g c) 1 mole Mg(OH)2 → 24 + 2( 16 + 1) = 58g 0.125 mole Mg(OH)2 → 0.125 x 58g = 7.25g
Conversion from Moles to Mass
Moles
Mass
Conversion from Moles to Volume
Volume
Calculate the volume of 0.75 mole of nitrogen at stp ( 1 mole occupies 22.4dm3 at stp)
Answer: a) 1 mole occupies → 22.4dm3
0.75 mole → 0.75 x 22.4dm3 = 16.8dm3
Calculate the moles for a) 23.5g of copper(II)nitrate, Cu(NO3)2
b) 0.97g of caffeine C8H10N4O2 molecules
Answer:
a) 1 mole copper(II)nitrate, Cu(NO3)2 → 64 + 2 [14 + 3(16)] = 188g 188g Cu(NO3)2 → 1 mole 23.5g Cu(NO3)2 → 1 x 23.5 = 0.125 mol 188 b) 1 mole, C8H10N4O2 → 8(12) + 10 + 4(14) + 2(16) = 194g 194g C8H10N4O2 → 1 mole 0.97g C8H10N4O2 → 1 x 0.97 = 0.005 mol 194
Mass
Moles Volume
Conversion from Mass to Volume
Calculate the volume of 3.4g of NH3 at stp ( 1 mole occupies 22.4dm3 at stp)
Answer:
a) 17g NH3 → 1 mole 3.4g NH3 → 1 x 3.4 = 0.2 mol 17 1 mol NH3 → 22.4dm3
0.2 mol NH3 → 22.4 x 0.2 dm3
= 4.48 dm3
Conversion from Mass to Moles
Calculate the number moles of gas at stp a) 4.8dm3 chlorine gas b) 1200cm3 methane gas
Answer: a) 22.4dm3 → 1 mole gas 4.8dm3 → 4.8 x 1 22.4 = 0.214 mol b) 22400cm3 → 1 mole gas 1200cm3 → 1200 x 1 224000 = 0.0535mol
Conversion from Volume to Moles Conversion from Volume to Mass
Calculate the mass of the following gas at stp ( 1 mole occupies 22.4dm3 at stp) a) 7.2dm3 of NH3 gas b) 600 cm3 of CH4 gas
Answer: a) 22.4dm3 NH3 → 1 mole gas 7.2dm3 NH3 → 1 x 7.2 22.4 = 0.32mol 1 mol NH3 → 17g 0.32 mol NH3 → 17 x 0.32g = 5.44g b) 22400cm3 CH4 → 1 mole 600cm3 CH4 → 1 x 600 22400 = 0.0267 mol 1 mol CH4 → 16g 0.0267 mol CH4 → 16 x 0.0267 = 0.4272g
Volume
Mole Mass
Calculate the number of particles in a) 12.8g of copper atoms, Cu b) 8.5g of ammonia NH3 molecules
Answer: a) 64g Cu atoms → 1 mole 12.8g Cu atoms → 1 x 12.8 = 0.2 mol 64 1 mol Cu → 6 x 1021 Cu atoms 0.2 mol Cu → 6 x 1021 x 0.2 = 1.2 x 1021 Cu atoms b) 17g NH3 molecules → 1 mole 8.5g NH3 molecules → 1 x 8.5 = 0.5 mol 17 1 mole NH3 → 6 x 1023 NH3 molecules 0.5 mole NH3 → 0.5 x 6 x 1023 = 3 x 1023 NH3 molecules
Conversion from Mass to Number of particles
Mass
Number particles
Moles
Conversion from Volume to Number of particles
Calculate number of particles at stp a) 0.28dm3 of N2 gas
Answer: a) 22.4dm3 N2 → 1 mole 0.28dm3 N2 → 1 x 0.28 22.4 = 0.0125mol 1 mol N2 → 6.02 x 1023 particles 0.0125 mol N2 → 6.02 x 1023 x 0.0125 = 7.5 x 1021 particles
Moles Volume
Number particles
Number particles
Mass Volume
Calculate the mass in a) 1.2 x 1022 zinc atoms
b) 3 x 1023 ethanol, C2H5OH molecules
Answer: a) 6 x 1023 Zn atoms → 1 mole 1.2 x 1022 Zn atoms → 1.2 x 1022 = 0.02 mol 6 x 1023 1 mol Zn atoms → 65g o.o2 mol Zn atoms → 0.02 x 65g = 1.3g
b) 6 x 1023 C2H5OH molecules → 1 mole 3 x 1023 C2H5OH molecules → 3 x 1023 = 0.5mol 6 x 1023
1 mole C2H5OH → 46g 0.5 mole C2H5OH → 46 x 0.5 = 23g
Mass
Number particles
Conversion from Number of particles to Mass
Moles
Conversion from Number of particles to Volume
Calculate volume of 1.5 x 1023 molecules of N2 at stp ( 1 mole occupies 22.4dm3 at stp)
Answer: a) 6.02 x 1023 N2 molecules → 1 mole 1.5 x 1023 N2 molecules → 1 x 1.5 x 1023
6.02 x 1023
= 0.25mol 1 mol N2 → 22.4dm3
0.25 mol N2 → 22.4 x 0.25 = 5.6dm3
Number particles
Volume Moles
Number particles
Mass Volume
Ideal Gas Law Lab simulation
Ideal Gas Law Videos