Week 8 – SS1 Second Term Chemistry Notes

 
 WEEK 8
GAS LAWS
Behaviour of gases is expected to differ from that of solids and liquids. This was investigated by many early scientists e.g. Boyle, Charles, Graham and Dalton, Avogadro’s. They studied the physical behaviour of gases. Gay-Lussac: he studied the chemical behaviour.
Boyle’s Law

• By Robert Boyle in 1662
• States that the volume of a given mass of a gas is inversely proportional to its pressure, provided temperature remains constant.
• Law is about the relationship between volume (V) and pressure (P)

As volume of a gas increases, the pressure decreases and vice versa
Mathematical expression
V α 1/p
V α k/p
K = PV
V = volume
P = pressure
K = constant
For initial and final pressure and volume, we have
P₁V₁ = P₂V₂
According to the kinetic theory,the gas pressure is caused by molecular collisions with the walls of the container.Therefore, the larger the number of molecules per unit volume,the larger the number of collisions and the higher the pressure.Reducing the volume of a gas container will increase the collisions on the walls of the container per unit time and consequently the pressure of the gas will incr

 
 
 
 Using kinetic theory to explain Boyle’s Law

 
NOTE:
In stage 2:

• The piston is kept stationary by placing a heavy weight on it.
  
• The space occupied by the molecules is constant i.e. volume is constant.
  
• The gas exerts constant pressure.
  

In stage II:

• Weight of piston is replaced with a lighter one, so the piston moves up.
  
• The space occupied by the gas is doubled (increased volume).
  
• The pressure exerted by the gas is halved (pressure increases).
  

 
 
 In stage III:

• Weight of the piston is now replaced with a heavier one, so the piston moved down.
• The space occupied by the gas is halved (volume decreases)
• The pressure exerted by the gas is doubled (pressure increases)

Therefore, at constant temperature; as the volume of a gas decreases, the pressure the gas exerts increases. Example, Ababio page 31, Fig. 5.15
V                        

 
  0                ¹/_P                            

Calculation Involving Boyle’s Law
The volume of a gas means nothing unless the conditions under which it was measured are known.
  Tips for working with gas laws:

• All gas calculations must use Kelvin temperatures. 
• The conditions 0 ^oC and 1 atm are referred to as standard temperature and pressure – STP. 
• The volume occupied by one mole of a gas at STP, 22.4 liters, is referred to as molar volume. 
• Read the problem to see what conditions change.  
• Decide which gas law to use and write its equation.  
• Reread the problem to see what question is asked.  
• If needed, manipulate the gas law equation.  
• Plug numbers and units into the equation.  
• Pickup your calculator and punch buttons.  
• Write the answer to the problem, don’t forget significant figures, and circle it.

  Example 1
A sample of gas occupied 390cm3 ata pressure of 760mmHg.What volume will the gas occupy at 780mmHg, if the temperature remains constant?
Solution
P₁V₁=P₂V₂ (T constant)
P₁=760mmHg
V₁=390cm³
P₂=789mmHg
V₂=?
P₁V₁=P₂V₂ (Boyle’s law)
V2=P₁V₁/P₂
=760mmHg x 390cm³/780mmHg
V2=380 cm³

1. 375cm³ of gas has a pressure of 770mmHg, Find its volume if the pressure is reduced to 750mmHg

   Ans = 385cm³
   
    
    
    
    
    
    ABSOLUTE TEMPERATURE
   The temperature at which the volume of a gas would be theoretically reduced to O. This temperature is 0C or 273K.
   Note: practically, it is not possible as all gases liquefy above this temperature but it is significant because it is the lowest possible temperature that can be reached.
   Temperature Conversion
   0⁰C =273K, -273⁰C =0K (no degree sign)
   C =Celsius or centigrade
   K = Kelvin

     To convert:
   Celsius to Kelvin =K=⁰C + 273
   Kelvin to Celsius = ⁰C = K-273

                            

         100⁰C     373K

    
                                
    0⁰C                    273K

     -273^OC 0K                
   CELSIUS SCALE                    KELVIN SCALE

    

CHARLE’S LAW    
The volume of a given mass of a gas is directly proportional to its absolute temperature provided that pressure remains constant.
MATHEMATICAL EXPRESSION

 
 
 V α T
V α KT
K= V
T
V = Volume, T = Temperature (in Kelvin), K= constant
For more than one gas, we have
V_{1 =}V₂
T₁ T₂
V₂₌ V₁ T₂
T₁
The volume of a gas is zero at a temperature of -273⁰C which is zero Kelvin.
Charle’s laws explains that the behavior of gases at differenttemperature changes when the pressure is constant.Gases expand when heated.The rate of expansion or contraction is summaried as follows.At constant pressure,a gas increases by 1/273 of its volume for each Celsius degree rise in temperature and this is true for all gases.For every one degrr centrigrade rise or fall in temperature.

 The Kelvin temperature scale has -273oC as its starting point and it is called the absolute temperature scale.
Hence,O⁰C=273K
-273=0
The Kelvin temperature scale has -273oC as its starting point and it is called the absolute temperature

 
 
 Charles’ Law Problems
Example1
Whatvolume would be occupied by a given sample of gas at 45⁰C if it occupies 500cm³ at O⁰c assuming the pressure is constant?
Solution
Acertain mass of gas occupies300cm3 at 35⁰C .At what temperature willit have its volume reduced by half,assuming the pressure remains constant.
Solution
V1/T1=V2/T2
V1=300cm³
T1=35⁰C=(273+35)K=308K
V2=150cm³
T2=?
300cm³/308K=150cm³/T2
T2=308K X 150cm³/300cm³
154K =(-119⁰C)
 
 
     

 
 
 
 
 
 
 
 _{USING KINETIC THEORY TO EXPLAIN CHARLE’S LAW. EXAMPLE.}

NOTE:
In stage:
The gas is heated, molecules acquire more kinetic energy, move faster and collide more often with the walls of the vessel, and hence, pressure is exerted.
In stage II:
The temperature is increased through heating (T₁ –T₂).
The pressure is still constant because the piston has been moved up but the volume occupied by the gas has increased.
Conclusion: As temperature increases, the volume increases and vice versa.
    
    

 
 
 
 Graphical illustration of Charles law.

Ababio page 81, fig 5.16 and page 83
Vol (dm³) Vol


-273 O Temp^oC K
    Calculations involving Charles law

1. At 17⁰C, a sample of hydrogen gas occupies 125cm³. What will the volume be at 100⁰C, if the pressure remains constant? Ans. = 161cm³
2. To what temperature in Celsius must a gas be raised from 0⁰C in order to double its volume? Ans. = 273⁰C
3. 20cm³ of a gas at 55⁰C exerts 160mm Hg pressure. At the same pressure, calculate the volume of the temperature is doubled? Ans. = 23.35cm³

   GENERAL GAS EQUATION
   Boyle’s and Charles’s laws show that there is a relationship between the temperature, press and volume. The relationship is expressed by what is called general gas equation I.e.
   Boyle’s law – V α 1/P
   Charles Law – V α T    i
   Multiply (i) x (ii)
   V α 1 X Vα T
   V α (1/P X T)
   V α T/P
   V= KT/P
   PV= KT
   K= PV/T

    
    
    
    
    
    
    
    
    Formorethan one gas,
   P₁ V₁ = P₂ V₂V₂= P₁ V₁ T ₂
   T₁    T₂ P₂ T ₁

where P1= initial pressure
V1=initial volume
T1=    Initial Kelvin temperature
P2=New pressure
V2=New volume
T2=New Kelvin temperature

 STANDARD TEMPERATURE AND PRESSURE (S.T.P)
It is the generally accepted standard temperature (0⁰C or 273K) and pressure (760mmHg or 1.01 x 10⁵Nm^-2)
NOTE: if two chemists, one in a temperate country e.g. England and the other in tropical country e.g. Nigeria, should carry out investigations on the same gases. Their gas volumes would differ because of different in temperatures of the two countries. So, scientists decided to have standard temperature and pressure for calculations and experimentation.
CALCULATIONS

1. At s.t.p, a certain mass of gas occupies a volume of 790cm³. Find the temperature at which the gas occupies 1000cm³. Ans. = 330.1K
2. A given mass of a gas occupies 850cm³ at 320K and 0.92 x 10⁵Nm^-2 pressures. Calculate the volume of the gas at s.t.p Ans. = 660.5cm³
3. A sample of N₂ occupies a volume of 1dm³ at 500k and 1.01 x 10⁵ Nm^-2. What will its volume be at 2.02 x 10⁵Nm^-2 and 400K?
4. To what temperature must a given mass of N₂ at 0⁰C be heated so that both its volume and pressure will be doubled?
5. 130cm³ of gas at 20^oC exerts a pressure of 750mmHg. Calculate its pressure if its volume is increased to 150cm³ at 35⁰C.
6. Calculate the volume of hydrogen produced at s.t.p and r.t.p when 25g of zinc are added to excess dilute Hydrochloric acid at 31^oCand 778mmHg pressure     (H= 1, Zn= 65, Cl= 35.5, GMV= 22.4dm^3, 24 dm³ )
7. The combustion of butane in oxygen (air) is represented in the equation below: 2C₄H₁₀ + 130₂ 10H₂0 + 8CO₂

    
    
    
    IDEAL GAS EQUATION
   In all experimental works, measurements or calculation involving gases, four quantities are important-volume, pressure, temperature, and number of moles. The first three have been used in general gas equation. But the combination of four of them gives ideal gas equation as:
   PV = nrt
   Where
   P = Pressure (in atm)
   V = Volume (in dm³)
   n = No. of moles
   R = Gas constant (0.082 atm dm³ k^-1 mol^-1)
   T = Temp (in K)

Gas density can also be calculated using the ideal-gas equation.
Density is equal to mass divided by volume, d = m/v.
The ideal-gas equation can be arranged to give density in g/L:

This equations shows the density of a gas depends on its pressure, molar mass, and temperature. The higher the molar mass and pressure, the greater the gas density; the higher the temperature, the less dense the gas.
Even though gases form homogeneous mixtures regardless of their identities, a less dense gas will lie above a more dense one if they are not physically mixed. The differences between the densities of hot and cold gases is responsible for CO₂ being able to keep oxygen from reaching combustible materials (thus acting as a fire extinguisher) and for many weather phenomena, such as the formation of large thunderhead clouds during thunderstorms.
 
 
 

1. Under a pressure of 3000, Nm^-2 a gas has a volume of 250cm^3. What will its volume be if the pressure is changed to 100mmHgat the same

   Temperature?
       Solution to NO.3

Note: The pressure is not in the same unit. So conversion must be done first. 101325 Nm^-2 = 760mmHg
3000Nm^-2 =760     x    3000= 22.gmmHg
101325
P_{1 =}22.5mmHg
Using         P₁ V_{1         =} P₂ V₂
        T₁         T₂
P_{2 =} 100mmHg, V₁ = 250cm^3, V_{2 =}?
Ans= 56.25cm³
Dalton’s Law of Partial Pressures, established by John Dalton, states that if there is a mixture of gases that do not react chemically together, then the total pressure exerted by the mixture is the sum of the partial pressures of the individual gases that make up the mixture i.e.
P_{Total =}P_A + P_B + P_C +…
Where

 
 P_{Total =} Total pressure of the mixture
P_{A =} Partial pressure of gas A
P_B = Partial pressure of gas B
P_C = Partial pressure of gas C
Note:

1. Gases A, B and C make up the mixture.
2. If a gas is collected by water, it is likely to be saturated with water vapor and the total pressure becomes: P_total= P_gas + P_watervapour

The pressure of the dry gas will now be:
P_{GAS =} P_TOTAL -P_WATER

 Dalton’s Law is helpful when collecting a gas “over water”. This diagram shows the collection of a gas by water displacement.
A collecting tube is filled with water and inverted in an open pan of water. Gas is then allowed to rise into the tube, displacing the water. By raising or lowering the collecting tube until the water levels inside and outside the tube are the same, the pressure inside the tube is exactly that of the atmospheric pressure.
A gas collected “over water” is a mixture of the gas and water vapor. Dalton’s law of partial pressures describes this situation as:

 
 
 
 
 P_total = P_gas + P_H2O
Charts like this one are readily available that give water vapor pressure at any common temperature.
EXERCISES

1. A certain mass of hydrogen gas collected over water of 6^o C and 765mmHg pressure has a volume of 35cm^3. Calculate the volume when it is dry at s.t.p. (s.v.p. of water at 6^oC=7mmHg).
   

   Solution
   To get the real pressure of H₂ gas i.e. when it is dry:
   P_{DRY GAS} =P_TOTAL –P_{WATER VAPOUR =}765 -7 =758mmHg
   Apply general gas equation to obtain the volume required in the question. Ans= 34.2cm³
   

2 .272cm³ of CO₂ were collected over water at 15^oC and 782mmHg pressure. Calculate the     Volume of the dry gas at s. t. p. (s.v.p of water at 15 0C= 12mmHg)     
3 .A given amount of gas was collected over water at 302K where the water vapour pressure of was 4.0 KNm^-2. Calculate the pressure of the dry gas if the atmospheric pressure at the same temperature was 101.3 KNm^-2
4 .200cm^-2 of Nitrogen gas at a pressure of 500mmHg and 100cm³ of CO₂ at a pressure of 50mmHg were introduced into a 150cm³ vessels. What is the total pressure in the vessel?
Solution for No. 4
For N₂
V₁ =200 cm³, P₁ =500 mmHg

 
 
  V2=150 cm³, P₂ =? mmHg
P₂= P₁V₁ = 666.67mmHg –Boyles Law
    V₂
For CO₂
V₁ =10 cm³, P₁=50 mmHg, V₂=150 cm³, P₂ mmHg=?
P₂-33.33mmHg.
Total pressure =666.67 +33.33 =700mmHg

1. A gas X was put in a 10dm³ vessel at 400K and 1.015 x 10⁵ Nm^-2. . Another gas Y at 400K and 3.035 x 105 Nm^-2 is put in a 5dm3 vessels. What will be the total pressure if:
   1. X and Y are put in a 5dm³ vessel at 400K?
   2. X and Y are put in a 15dm³ vessel at 400K?
   3. X and Y are put in a 10dm³ vessel at 400K?
   4. X and Y are put in a 20dm³ vessel at 400K?

      SOLUTION
      For X

      1. P₁= 1.015 x 105 Nm^-2, V₁ = 10dm³, V₂ = 5dm³, P₂=?

         P₂ = P₁V₁    =2.03 X 105 Nm^-2.         NOTE: Temp is constant
         V₂

    For Y
It already in 5 dm³ and it is 3.035 x 10⁵ Nm^-2
P _total = Px + Py
=2.03 x 10⁵ + 3.035 10⁵
=5.065 x 10⁵ Nm^-2
EVALUATION
EXERCISES
1.A certain amount of gas occupies 5.0dm³ at 2 atm and 10 ⁰ C. Calculate the number of moles present (R=0.082atm dm ³ K^{– 1}mol- ¹⁾.

1. 2.0 moles of an Ideal Gas are at a temperature of -13⁰C and a pressure of 2

 
 Atm. What volume in dm³ will the gas occupy at a temperature?
(R=0.082 atm dm³ K- ¹ mol-¹)

 3.A given mass of nitrogen is 0.12dm³ at 60C and 1.01 X 10⁵ Nm^-2. Find its pressure at the same temperature, if its volume is changed to 0.24dm^3.
4.A certain mass of gas occupies 600cm³ and exerted 1.325 x 10⁵ Nm^-2 pressures. At what pressure would the volume of the gas be halved?
Ans = 2.65 x 10⁵ Nm^-2

 5.Convert the following Kelvin temperature to Celsius temperature
A. 405K B. 298K

 6.Convert the following Celsius temperature to Kelvin temperature
A. 0⁰C B. -132⁰C