Categories
Economics

General gas laws

GENERAL GAS LAW

From the gas laws, we know that the volume of a gas depends on both its temperature and pressure. The relationship between the three variable; i.e. volume,, temperature and pressure can be summarized up as follows:

If V∝1P (Boyle’s law at constant temperature) and V∝T (Charle’s law at constant pressure)

V∝1P×T(both temperature and pressure may vary) or PVT=K (a mathematical constant for a fixed mass of gas)

PVT=K is often known as the general gas equation.

GENERAL GAS EQUATION

General gas equation states that for fixed mass of a gas under any set of conditions of V, P and T, the value of PVT  must remain constant. If for a fixed mass of gas V1 is the volume at pressure

P1 and absolute temperature T1 and V2 is the volume at pressure P2 and absolute temperature T2 it follows that.

P1V1T=P2V2T2

The general gas equation can be used to find the volume of a gas when both its pressure and temperature change. Thus;

V2=P1V1T2P2V2

The standard temperature and pressure

The value of gases are sometimes given in standard temperature and pressure (S. T. P). These values are standard temperature= 273k and standard pressure = 760mmHg. The S.I unit of standard pressure when used is 1.01 × 103Nm-2

Examples

  1. At S. T. P a certain mass of gas occupies a volume of 790cm3, find the temperature at which the gas occupies 1000cm3 and has a presence of 720mmHg

P1V1T1=P2V2T2

P1 = 760mmHg (at stp),     V1= 790cm3

T= 273k (at stp),  = 1000cm3

P= 726mmHg

T= New Temperature

∴T2=P2V2T1P1V1=720×1000×273760×790=330.1k

  1. A given mass of gas occupies 850cm3at 320k and 0.92 × 103Nm-2 of pressure. Calculate the volume of the gas at S.T.P.

P1V1T1=P2V2T2

P= 0.92 × 103Nm-2          T= 320k

V= 850cm3             P= SP + 1.01 × 103Nm-2

T= 273k (at stp)

V2 = new volume of gas.

∴V2=P1V1T2P2T1=0.92×850×2731.01×103×320=660.5cm3

EVALUATION

  1. Explain the general gas equation.
  2. If the volume of a given mass of a gas at 298k and a pressure of 205.2 × is 2.12 . What is the volume of the gas S.T.P (standard pressure= 1013 105Nm-2, standard temperature = 273)

IDEAL GAS LAW

The ideal gas: This is a gas sample whose properties correspond, within experimental error, to the relationship PV =RT. An ideal gas must obey all the rules guiding Boyle’s and Charles’s laws. Ideal gas conforms to the kinetic theory of gases. Four quantities’ are important in all experimental work, measurements or calculations involving gases. They are:

  1. volume
  2. pressure
  3. temperature and
  4. numbers of moles

Ideal gas equation is given by PV=nRT

The value of R for one mole of a gas at 273K, 1atm and volume 22.4dmis 0.0821atmdm3K-1mol-1 or 8.314JK-1mol-1

Examples:

  1. Calculate the volume occupied by 2.5moles of an ideal gas at -23oC and 4.0atm. (R = 0.0821atmdm3K-1mol-1)

Solution:

Using PV = nRT

where P = 4.0atm

n = 2.5 mole

T = −23 + 273 = 250K

Hence, V=nRTP=2.5×0.0821×2504=12.8dm3

NOTE:  Pressure can also be measured in other units. 760mmHg = 1atm = 101325Nm-2

Ideal gases only exist at experimental conditions of high pressure and low temperature. Basically all gases are real

REASONS WHY REAL GASES DEVIATE FROM IDEAL GAS BEHAVIOUR

  1. The forces of attraction in real gases are not negligible.
  2. The volume of real gases are not negligible. Hence, real gases have their own volume called excluded volume.
  3. Real gases undergo inelastic collision

EVALUATION:

  1. What is an ideal gas?
  2. Write down the ideal gas equation for n-mole of a gas.

GAY- LUSSAC’S LAW AND AVOGADRO’S LAW

Gay- Lussac’s law describes the combining volumes of gases that react together. In his experiment, all temperatures and pressures were kept constant:

  1. STEAM:Gay- Lussac’s observed that two volumes of hydrogen reacted with one volume of oxygen to yield two volumes of steam
2H2(g)+O2(g)2H2O(g)
Volume2 :12
Ratio2 :12
  1. HYDROGEN CHLORIDE GAS:One volume of hydrogen combined with one volume of chloride to yield two volumes of hydrogen.
Hydrogen+ChlorineHydrogen Chloride
i.eH2(g)+Cl2(g)2HCl(g)
Volume1:12
Ratio1:12

C.

Carbon (II) oxide+OxygenCarbon (IV) Oxide
2CO(g)+O2(g)2CO2(g)
Ratio2 :12

Gay- Lussac’s noticed that the combining volumes as well as the volumes of the products, if gaseous, were related by simple ratios of whole numbers. He proposed the law of combining volume or gaseous volumes.

Hence; Gay- Lussac’s law combining volumes states that when gases react, they do so in volumes which are in simple ratios to one another and to the volumes of the products, if gaseous provide that the temperature and the pressure remain constant.

EXAMPLES

  1. What is the volume of oxygen required to burn completely 45 of methane (CH4)?

Equation of reaction

CH4(g)+2O2(g)CO2(g)+2HO2(g)
Volume1:212
Ratio1:212

By Gay- Lussac’s Law:

1 volume of methane required 2 volumes of oxygen i.e.

1cm3 of methane requires 2cm3 of oxygen

∴ 45cm3 of methane require 90cm3 of oxygen

  1. 20cm3of carbon (I) oxide are sparked with 20cm3 of oxygen. If all the volumes of gases are measured at S.T.P, calculate the volume of the residual gases after sparking.
Equation of reaction2CO(g)+O2(g)2CO2(g)
Combining volume2:1:2
Volumes before sparking20cm310cm320cm3
Volumes after sparking-1020
Residual gases =un-reacted oxygen+Carbon (IV) oxide formed

Volume of residual gas = 10cm+ 20cm3 = 30cm3

AVOGADRO’S LAW

Avogadro’s Law states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

This law means that for all of gases e.g. oxygen, hydrogen, Chlorine etc if their volumes are the same, they will have the same number of molecules.

Avogadro’s Law is easily applied to convert volume of gases to the number of molecules. Avogadro’s Law can be used to solve problem under Gay –Lussac’s law of combining volumes.

The formation of steam from reaction of Hydrogen and Oxygen is given below:

ReactionHydrogen+OxygenSteam
Volume212
Gay-Lussac’s:2:1:2
Avogadro’s Law2:1:2

This agrees with the equation below:

2H2(g) + O2(s)  →   2H2O(s)

i.e 2 molecules of hydrogen combine with 1 molecule of oxygen to produce 2 molecules of steam

Example:

  1. 60of hydrogen are sparked with 20 of oxygen at 1000C and 1 atmosphere. What is the volume of the steam produced?

Solution

2H2  + O2   →   2H2O

From the equation, 2 molecules of hydrogen react with 1 molecule of oxygen to produce 2 molecules of steam.

2H2+O22H2O
2vols1vol2vols (combining volumes)
i.e. 2cm31cm32cm3

From the above information, when 2cm3(2 vol) of H2 react, 1cm3(1 vol) of Owill react i.e. half of H2vol, to give 2cm3(2 vol) of H2O.

Thus, 10cm3 of H2 will react with 5cm3 of O2 to produce10cm3 of H2O and so on.

From the question, we have 60cm3 of H2 and 20cm3 of O2, thus, when all the 20cm3 of O2 react, only 40cm3 of H2 will react to give 40cm3 of H2O, because the volume of H2 is the same as that of H2O i.e.

2H2+O22H2O
2vols1vol2vols
2cm31cm32cm3
40cm320cm340cm3

Thus, the volume of steam (H2O) formed is 40

  1. What volume of propane is left unreacted when 80 of oxygen and 20 of propane react according to the equation below?

C3H8(g)   + 5O2(g)   →     3CO2(g) + 4H2O

C3H8(g)+5O2(g)3CO2(g)+4H2O
1vol5vols
1cm35cm3
4cm320cm3

Volume of the propane before the reaction = 20cm3

The volume that reacted = 4cm3

Volume that did not react = volume before reaction – volume that reacted i.e. 20 – 4 = 16cm3

EVALUATION

  1. State Gay–Lussac’s law
  2. State Avogadro’s law.
  3. 50cmof methane were burnt completely in oxygen according to the equation below.

CH4 + 2O2    →    Co2 + 2H2O

Calculate: (a) volume of oxygen used (b) volume of carbon(Iv) oxide produced (c) volume of steam produced.

GRAHAM’S LAW OF DIFFUSION.

This law states thatat constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass or square root of its vapour density.                  

Mathematically, Graham’s law of diffusion can be represented as: R1R2∝P2P1−−√

Where Rand R2 are the rates of diffusion and P1 and P2, the densities of the two gases.

The density is directly proportional to its molecular mass.

EXAMPLES

  1. 100cm3of oxygen diffuse through an office in 60 seconds while it takes120 seconds for the same office. Calculate the molecular mass of the unknown gas [0 = 16]

Solution

RRxO2=MxMO2

Since the rate of diffusion is inversely proportional to the time taken:

RRxO2=txtO2=MxMO2−−−−√(txtO2)=MxMO2Mx=MO2×(txtO2)2=32×(12060)2=32×22Mx=32×4=128g

200cm3 of hydrogen diffused through a porous pot in 40 seconds. How long will it take 300cm3 of chlorine to diffuse through the same pot?

Solution

200cm3 of hydrogen diffused in 40secs

∴ 300cm3 of chlorine will diffuse in

300cm3200cm3×40(3×20)=60seconds

Now, using the equation,

t1t2=M1M2−−−√

Where  t = 60s,

M1 = molecular mass of hydrogen

i.e H2=(2×1)=2

M2 = molecular mass of chlorine  =cl2=2×35.5=71

T2=t1M1M2−−−√=60712−−√=6035.5−−−−√=60×5.96=357.5sec

Time of diffusion of chlorine = 358s.

How many times the rate of diffusion of hydrogen is faster than that of oxygen and what law do you use to get the answer? [vapour density] of [H=1, O=16]

Solution

Rate (R+) of diffusion of H2

=Density of O2Density of H2−−−−−−−−−√R1R2=161−−√=41

∴ Hydrogen diffuses four times faster. The law used is Graham’s law of diffusion.

The vapour density of a gas or vapour is the number of times a given volume of gas (or vapour) is heavier than the same volume of hydrogen measured and weighed under the same temperature and pressure.

Vapour density =mass of 1 vol of a gas or vapourmass of equal volume of hydrogen

Applying Avogadro’s law, it is possible to show that the vapour density of a gas is related to the relative molecular mass of the gas.

V.D.=mass of 1 mole of a gas or vapourmass of 1 molecule of hydrogenV.D.=mass of 1 vol of a gasmass of 2 atoms of hydrogen

∴2×V.D.= relative molecular mass

The density of hydrogen at S.T.P is 0.09cm3

Example

Calculate the vapour densities of the following gases from the given data.

  1. 560cm3of oxygen at S.T.P weighs 0.8g
  2. 1,400cm3of sulphur (iv) oxide weighs 4g

Solution

  1. 1000cmof hydrogen at S.T.P weighs 0.09g

∴ 560cm3 of hydrogen at 560cm3100cm3×0.09=0.05g

V.D.=mass of a given volume of gasmass of equal volume of hydrogen

∴ Vapour density of oxygen =mass of 560cm3of oxygenmass of 560cm3of hydrogen

  1. 1000cm3of hydrogen at S.T.P weighs 0.09g.

∴ 1400 of hydrogen will weigh 1400×0.091000=0.126g

Vapour density =mass of a given volume of gasmass of equal volume of hydrogen

∴ Vapour density of SO=mass of 1400cm3 of SO2mass of 1400cm3 of H2=4g0.126=31.74≅32

EVALUATION

  1. Deduce the relationship between relative molecular mass and vapour density of a substance.
  2. Define vapour density of a gas.

MOLAR VOLUME OF GASES- AVOGADRO NUMBER AND THE MOLE CONCEPT

The molar volume of any gas is the volume occupied by one mole of that gas at s.t.p. and is  numerically equally to 22.4dm3  i.e. one mole of any gas at s.t.p. occupies the same volume the value of which is 22.4dm3 . This value is called molecular mass or molar mass.

From the Avogadro’s law, the molar volume for all gases contains the same number of molecules. This number is called the Avogadro’s number or constant and the value is 6.02×1023 at s.t.p

MOLE: The mole can be defined as the amount of substance which contain as many elementary particles or entities e.g. ions, molecules, atoms, electrons as the number of atoms in exactly 12 grams of carbon -12.

The mole of any substance represents 6.02×1023 particles of any substance. Therefore, a mole refers to Avogadro’s number of particles of any substance.

In summary, the molar mass of a gas contains Avogadro’s number of molecules which is 6.02×1023 and occupies a volume of 22.4dm3  at s.t.p.

The atomic mass of every element also contains Avogadro’s number of atoms.

The mole concept– This says that one mole of any substance contains the same number of particles; which can be atoms, molecules or ions. This number is 6.023×1023dm3 (the Avogadro’s number)

Examples

  1. 158g of a gas at s.t.p. occupies a volume of 5000dm3. What is the relative molecular masss of the gas? (Molar volume at s.t.p= 22.4dm3mol-1

Solution

Volume of gas: V = 50.00dm3

Molar volume of gas; V = 22.4dm3 mol-1

N = amount in moles

=vvN=5022.4dm3mol−1=2.23mol

Molar mass M of the gas =Mn=158g22.4dm3mol−1=70.8

Molar mass = 71 gmol-1

  1. What is the mass of 3 moles of oxygen gas O2? (O = 16)

Mass of 1 mole of O2=(2×16)g=32g

Mass of 3moles of O2=(3×32)g=96g

  1. How Many moles are there in 20g of CaCO3? [CaCO3= 100]

Molar mass of CaCO= 100g

100g of CaCO= 1 mole

20g of CaCO=20100×1mole=0.2mole

EVALUATION

  1. Using the relationship between mole and Avogadro’s number. Define mole in six ways.

DALTON’S LAW OF PARTIAL PRESSURE

Dalton’s law of partial pressure states that for a mixture of gases that do not react chemically, the total pressure exerted by the mixture of gases is equal to the sum of the partial pressures of the individual gases.

Mathematically, Dalton’s law of partial pressure for a mixture of n gases can be expressed as:

Ptotal = P1 +P2+P3 +………..+ Pn where Ptotal is the total pressure exerted by the mixture of gases that dot not react, P1, P2, P3……Pn are partial pressure of the individual gases.

Example:

If 20.0dm3 of hydrogen were collected over water at 17oC and 79.7kNm-2 pressure; Calculate the

(a) Pressure of dry hydrogen at this temperature.

(b) Volume of dry hydrogen at s.t.p.

(vapour pressure of water is 1.90 kNm-2 at 17OC)

Solution:

(a)

PH2  =  Ptotal – Pwater vapour

=  79.7 – 1.90

=  77.8 kNm-2

(b)

P1V1T1=P2V2T277.9×20290=101.3V2273V2=14.5dm3

  1. (a) State Graham’s law of diffusion

(b) Arrange the following gases in decreasing order of diffusion rate: Chlorine, hydrogen chloride, hydrogen sulphide and Carbon(IV) oxide

[H = 1, C = 12, O = 16, S = 32, Cl = 35.5]

  1. (a)  What do you understand by s.t.p?

(b) If the volume of a given mass of gas at 298k and pressure of 205.2 × 103 Nm−2 is 2.12dm3, what is the volume at S.T.P? Standard pressure= 101.3 × 103 Nm. Standard temperature = 273k

  1. Calculate the number of moles of the following at s.t.p

(a) 16g of oxygen

(b) 67.2dm3 of nitrogen gas, and

(c) 1.14dm3  of hydrogen chloride gas.

O = 16, H = 14, N =1. Molar volume of gas at S.T.P = 22.4dm3

(b) (i) Convert 33ºC  and -41ºC  to Kelvin scale

(ii) Convert 270k and 315k to 0ºC

Join Discussion Forum and do your assignment: Find questions at the end of each lesson, Click here to discuss your answers in the forum

Ad: Get a FREE Bible: Find true peace. Click here to learn how you can get a FREE Bible.

For advert placement/partnership, write [email protected]

Download our free Android Mobile application: Save your data when you use our free app. Click picture to download. No subscription. stoplearn

We are interested in promoting FREE learning. Tell your friends about Stoplearn.com. Click the share button below!