- 1. http://lawrencekok.blogspot.com Prepared by Lawrence Kok
Tutorial on Molar Volume, Boyles, Charles, Avogadro Law and Ideal
Gas Equation.
2. Ideal Gas Equation PV = nRT 4 different variables P, V, n, T
T = Absolute Temperature in K R = universal gas constant Unit -
8.314Jmol-1 K-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 3.
Ideal Gas Equation PV = nRT (n, T fix) PV = constant V = constant/P
V 1/p Charless Law PV = nRT 4 different variables P, V, n, T
Avogadros Law T = Absolute Temperature in K 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 R = universal gas constant Unit - 8.314Jmol-1 K-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 Boyles Law Pressure Law PV = nRT (P, T fix) V =
constant x n V n 4. Ideal Gas Equation PV = nRT (n, T fix) PV =
constant V = constant/P V 1/p Charless Law PV = nRT 4 different
variables P, V, n, T Avogadros 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-1 K-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 Boyles Law Pressure Law P1 = P2 T1 T2 PV = nRT
(P, T fix) V = constant x n V n 5. PV = nRT (n, T fix) PV =
constant V = constant/P V 1/p Charless Law Avogadros 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 Boyles Law 6. PV
= nRT (n, T fix) PV = constant V = constant/P V 1/p Charless Law
Avogadros 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 Boyles Law PV = nRT (n, T fix) PV = constant V = constant P V
inversely proportional to P V 1 P P1V1 = P2V2 Boyles Law 7. PV =
nRT (n, T fix) PV = constant V = constant/P V 1/p Charless Law
Avogadros 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 Boyles Law PV = nRT (n, T fix) PV = constant V = constant P V
inversely proportional to P V 1 P P1V1 = P2V2 Boyles Law Lab
Simulator Video on Boyles Law Boyles Law 8. PV = nRT (n, T fix) PV
= constant V = constant/P V 1/p Charless Law Avogadros 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 Boyles Law 9. PV
= nRT (n, T fix) PV = constant V = constant/P V 1/p Charless Law
Avogadros 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 Boyles Law Charless 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 10. PV = nRT (n, T fix) PV = constant V =
constant/P V 1/p Charless Law Avogadros 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 Charless Law Lab Simulator
Video on Charless Law Boyles Law Charless 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 11. PV = nRT (n, T fix) PV =
constant V = constant/P V 1/p Charless Law Avogadros 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 Boyles Law 12. PV
= nRT (n, T fix) PV = constant V = constant/P V 1/p Charless Law
Avogadros 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 Boyles 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 13. PV = nRT (n, T fix) PV = constant V
= constant/P V 1/p Charless Law Avogadros 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 Boyles 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 14. PV = nRT (n, T fix) PV =
constant V = constant/P V 1/p Charless Law Avogadros 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 Boyles Law 15. PV
= nRT (n, T fix) PV = constant V = constant/P V 1/p Charless Law
Avogadros 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 Boyles Law Avogadro Law PV = nRT (P, T fix) V = constant x n
V directly proportional to n V n V1 = V2 n1 n2 16. PV = nRT (n, T
fix) PV = constant V = constant/P V 1/p Charless Law Avogadros 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 Boyles Law Avogadro Law PV
= nRT (P, T fix) V = constant x n V directly proportional to n V n
V1 = V2 n1 n2 17. 1 mole of any gas at fix STP (Std Temp/Pressure)
occupies a volume of 22.4dm3 /22400cm3 /24L
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 T 0C (273.15K)
Unit conversion 1 atm = 760 mmHg/Torr = 101 325Pa(Nm-2 )
=101.325kPa 1m3 = 103 dm3 = 106 cm3 1dm3 = 1 litre P - 1 atm = 760
mmHg = 101 325Pa (Nm-2 ) = 101.325kPa Standard Molar Volume
Standard Temp/Pressure 22.4L 22.4L 1 mole gas 18. 1 mole of any gas
at fix STP (Std Temp/Pressure) occupies a volume of 22.4dm3
/22400cm3 /24L Avogadros 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 = 106 cm3 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 Avogadros Law 1
mole gas 19. PV = nRT (n, T fix) PV = constant V = constant/P V 1/p
Charless Law Avogadros 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 Boyles Law Charless Law Boyles Law
Pressure Law Avogadros Law 2 different variables 20. PV = nRT (n, T
fix) PV = constant V = constant/P V 1/p Charless Law Avogadros 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 Boyles Law Combined Gas Law Boyles Law Charless 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 Charless Law
Boyles Law Pressure Law Avogadros Law 2 different variables 21. PV
= nRT (n, T fix) PV = constant V = constant/P V 1/p Charless Law
Avogadros 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 Boyles Law Combined Gas Law Boyles Law Charless
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
Charless Law Boyles Law Pressure Law Avogadros Law Combined Boyle
Law + Charles Law Combined Gas Law 2 different variables 2
different variables 3 different variables 22. PV = nRT (n, T fix)
PV = constant V = constant/P V 1/p Charless Law Avogadros 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
Boyles Law Charless Law Boyles Law Pressure Law Avogadros Law 2
different variables 23. PV = nRT (n, T fix) PV = constant V =
constant/P V 1/p Charless Law Avogadros 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 Boyles Law Boyles Law
Charless Law V 1 P V n Boyle + Charles + Avogadro Law
Proportionality constant Gas constant, R V n T P 4 different
variables Charless Law Boyles Law Pressure Law Avogadros Law 2
different variables PV = nRT Ideal Gas Equation Avogadros Law V T
PV = n R T 24. PV = nRT (n, T fix) PV = constant V = constant/P V
1/p Charless Law Avogadros 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 Boyles Law Boyles Law Charless Law
V 1 P V n Boyle + Charles + Avogadro Law Proportionality constant
Gas constant, R V n T P 4 different variables Charless Law Boyles
Law Pressure Law Avogadros Law Boyle + Charles + Avogadro Law Ideal
Gas Equation 2 different variables PV = nRT Ideal Gas Equation
Avogadros Law V T PV = n R T 25. Charless Law Pressure Law PV = nRT
Avogadros LawBoyles Law V 1 P V T P TV n 2 different variables 26.
Charless Law Pressure Law PV = nRT Avogadros LawBoyles Law V 1 P V
T P TV n For 1 mole PV = RT For n mole PV = nRTP1V 1 = P2V2 T1 T2
PV = nRT Ideal Gas EquationCombined Gas Law + + 2 different
variables 3 different variables 4 different variables 27. Charless
Law Pressure Law PV = nRT Avogadros LawBoyles Law V 1 P V T P TV n
When n = 1 mol Gas constant, R is 8.31 JK-1 mol-1 or NmK-1 For 1
mole PV = RT For n mole PV = nRTP1V 1 = P2V2 T1 T2 PV = nRT Ideal
Gas EquationCombined 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-1 mol-1 or NmK-1
Find R (Universal Gas Constant) at molar volume n = 1 mol T = 273K
P = 101325Pa/Nm-2 V = 22.4 x 10-3 m3 R = ? 28. 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-1 mol-1 or NmK-1 29. 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-1 mol-1 or 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-1 K-1 1 atm 760 mmHg/Torr 101 325Pa/Nm-2
101.325kPa 1m3 103 dm3 106 cm3 1dm3 1000cm3 1000ml 1 litre x 103 x
103 cm3 dm3 m3 x 10-3 x 10-3 Unit conversion Different Units Used
30. A gas occupy at (constant P) V - 125cm3 T - 27C Calculate its
volume at 35C Find final vol, V2, gas at (constant T) compressed to
P2 = 250kPa V1 - 100cm3 P1 - 100kPa V2 - ? P2 250kPa What volume
(dm3) of 1 mol gas at P - 101325Nm-2 T - 25C Find volume (m3) of 1
mol of gas at T - 298K P - 101 325Pa Find volume (dm3) of 2.00g CO
at T 20C P 6250Nm-2 IB Questions on Ideal Gas 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 1 2 3 4 5 6 31. 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 32. IB Questions on Ideal
Gas 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 Find volume fixed mass gas when its pressure and temp
are 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 Fix
mass ideal gas has a V1 = 800cm3 , P1, T1 Find vol, V2 when P and T
doubled. V2 = ? P2 = 2P1 T2 = 2T1 A. 200 cm3 B. 800 cm3 C. 1600 cm3
D. 3200 cm3 33. 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 34. 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 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 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 A. 10kPa B.
20kPa C. 40kPa D. 480kPa 13 35. 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 = 4P1V2 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
36. 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 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 What
conditions would one mole of, CH4, occupy the smallest volume? A.
273 K and 1.01105 Pa B. 273 K and 2.02105 Pa C. 546 K and 1.01105
Pa D. 546 K and 2.02105 Pa 14 15 16 37. 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 1.01105 Pa B. 273 K and 2.02105 Pa C. 546 K
and 1.01105 Pa D. 546 K and 2.02105 Pa 14 15 16 38. 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 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) 39.
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 40. Calculate
the moles for a) 23.5g of copper(II)nitrate, Cu(NO3)2 b) 0.97g of
caffeine C8H10N4O2 molecules 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) Conversion from Mass to Moles 41.
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 42. Calculate the
number moles of gas at stp a) 4.8dm3 chlorine gas b) 1200cm3
methane gas 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
Volume Mole Mass 43. 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
44. Calculate the number of particles in a) 12.8g of copper atoms,
Cu b) 8.5g of ammonia 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 MolesVolume Number particles Number particles Mass Volume
45. 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
MolesVolume Number particles Number particles Mass Volume 46.
Calculate the mass in a) 1.2 x 1022 zinc atoms b) 3 x 1023 ethanol,
C2H5OH molecules 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) Number particles Volume Moles Number
particles Mass Volume 47. 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 48. Ideal Gas Law Lab simulation Ideal Gas Law Videos
49. Acknowledgements Thanks to source of pictures and video used in
this presentation Thanks to Creative Commons for excellent
contribution on licenses http://creativecommons.org/licenses/
Prepared by Lawrence Kok Check out more video tutorials from my
site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com
