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