CONTENTS:

1. Boyle’s law 
2. Charles’ law
3. General gas law
4. Ideal gas law

 

PERIOD 1:  BOYLE’S LAW

The relationship between volume and pressure of a gas was first started by Robert Boyle in 1662.

Boyle’s law states that the volume of a given mass of gas is inversely proportional to its pressure, provided that the temperature remains constant.

According to Boyle’s law, volume of a gas increases as the pressure decreases and vice versa.

This relationship is independent of the nature of the gas and it can be expressed mathematically as:

V α P

∴ V = kp

Or PV =k

Where V= volume at pressure P

          K = a mathematical constant

For a given mass of a gas, the product of its pressure and its volume is always a constant. If the pressure of a given mass of gas increases, its volume will decrease by a similar proportion and vice versa, as long as the temperature remain constant. This relationship can also be expressed mathematically as:

P1V1 = P2 V2

Where V1 = volume at pressure P1

V2= Volume at pressure P2

Boyle’s law can still be re-stated as: ‘The pressure of a given mass of gas is inversely proportional to its volume, provided the temperature remains constant. Boyle’s law can further be illustrated with the diagram below, showing that when P is increasing, V is decreasing and when P is decreasing, V is increasing.

P1 = Initial pressure

P2= Final pressure

V1=Initial volume

V2= final volume

Graphical representation of Boyle’s law

 

Worked examples

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

P1V1 = P2V2(Boyle’s law)

P1 = 770mmHg

P2 = 750mmHg

V1=375cm3

V2 =?  (New volume of gas)

P1V1 = P2 V2

V2 = P1 V1P2 =770 375750 =385cm3

The new volume will be 385cm3

 

2. 100cm3 of a gas has pressure of 1 atmosphere. Determine the volume of the gas at 5 atmospheres keeping the temperature constant.

Solution: since T is constant, we are to use Boyle’s law.

P1 Initial pressure = l atmosphere

P1Final pressure= 5

V1 Initial volume= 100cm3

 V2 (required quantity)

Recall: V2 = P1V1 = P2V2

V2=P1 V1P2 =100 15 = 20cm3

 

EVALUATION:

1. A given quantity of gas occupies a volume of 228cm3 at a pressure of 750mmHg. What will be its at atmospheric pressure if temperature is kept constant?
2. A given mass of gas at 550C has a pressure of 3.6×104Nm-2 and occupies a volume of 1.8dm3. What volume will it occupy if its pressure is increased to 4.8×104Nm-2 if the temperature is kept constant?

 

PERIOD 2: CHARLES’ LAW

The effect of temperature changes on the volume of a given mass of a gas at a constant pressure is described by Charles. Charles’ law states that the volume of a given mass of gas is directly proportional to its temperature in Kelvin, provided that pressure remains constant.

The volume of the gas decreases as the temperature decreases, and increases as the temperature increases.

Mathematically, the law can be expressed as:

V T

∴ V =kT

Or VT = k

Where v= volume

           T= Kelvin Temperature

           K= mathematical constant

A Representation of Charles’s law

 

                             

      

 

For a direct relationship, when the temperature increases, the volume will also increase at the same rate and vice versa, at constant pressure. The diagram above shows that when V is decreasing, T is also decreasing and when V is increasing, T is also increasing thus, making the quotient constant.

Charles’s law can be represented graphically has shown below.

 

If we divide the varying gas volumes by the corresponding temperature in Kelvin, the result would always be a constant. This relationship can also be expressed in another form.

V1T1   = V2T2∴ V2=T2 V1T1

Where V1 is the volume at temperature T1

          V2 is the volume at temperature T2

ABSOLUTE ZERO

This is the temperature at which the volume of a gas is theoretically zero. At this temperature there is no motion of any form and all gases have been liquefied or solidified. The value of the temperature is -2730C.

TEMPERATURE CONVERSION

1. To convert from Celsius scale to Kelvin scale, add 273 i.e. T= 0C + 273. This is because O0C=273K.
2. To convert from Kelvin scale to Celsius scale, subtract 273. i.e.

0C= T- 273.

Where T= Temperature in Kevin

0C= Temperature in Celsius.

Examples:

1.Convert the following Celsius temperature to Kelvin temperature.

1000C (b) 00C (c) -570C

Solution

Recall: T= 0C + 273

1000C= (100 + 273) = 373k

0C=(0 + 273) = (0 + 273) = 373k

-570c = (- 57 + 273)k = (273-57)= 216k

2. Convert the following Kelvin temperatures to Celsius temperature.

298k (b) 405k (b) 285k (d) 0k

Solution

Recall 00c = k – 273

298k = (298 – 273)0C= 250C

405k = (405 – 273)0C = 120C

0k     = (0 – 273)0C = – 2730C

Worked examples on Charles’s law

1. A gas occupies a volume of 20.0dm3 at 373k. Its volume at 746k at that pressure will be? 

Here pressure is constant. Charles’s law will apply.

V1=20.0dm3

T1 = 273k

  T2= 746

V2= ?

Recall Charles’s law = V1T1 =V2T2V2= V1 T2T1V2 = 20 ×746273 = 40.0dm3

EVALUATION:

1. State Charles’s law
2. Express the two laws mathematically
3. Draw two graphs to illustrate Charles’ law.

PERIOD 3: 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) orPVT = 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 =  P1 V1T2  ÷P2 V2

The standard temperature and pressure

The value of gases is 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 

T1 = 273k (at stp), V2 = 1000cm3

P2 = 726mmHg

T2 = New Temperature

∴T2  =P2  V2 T1P1 V1

= 720 ×1000 ×273760 ×790 = 330.1k

The new temperature of the gas is 330.1k

2. 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

P1= 0.92 × 103Nm-2 T1= 320k

V1= 850cm3 P2= SP + 1.01 × 103Nm-2

T2= 273k (at stp)

V2 = new volume of gas.

∴V2 = P1 V1 T2P2 T1  = 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 ×105Nm-2 is 2.12dm3. What is the volume of the gas S.T.P (standard pressure= 1013 105Nm-2, standard temperature = 273)

 

 

4: 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. An ideal gas conforms to the kinetic theory of gases. Four quantities are important in all experimental work, measurements or calculations involving gases. They are: (i) volume (ii) pressure (iii) temperature and (iv) number 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 22.4dm3is 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   =  2.5X0.0821X2504    = 12.8dm3.

               = 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.

 

GENERAL EVALUATION

OBJECTIVE TEST

1. A gas occupies 30.0dm3 at S.T.P. What volume will occupy at 910C and 52662.5Nm-2. (a) 20.0dm3 (b) 40.0dm3 (c) 60.0dm3 (d) 76.96dm3
2. Gases can be easily compressed because. (a) The molecule are relatively far apart (b) the molecules are quite close together (c) the molecule are very soft (d) the molecules are in constant, rapid motion
3. A give mass of gas occupies X1cm3 at Y1K. When the temperature is changed to Y2K, the volume becomes X2cm3, the pressure remaining constant. Which of the following equations correctly express the relationship between X1X2Y1andY2? (a) X1Y1 = X2Y2(b)X1Y1 = X2Y2 (c) X1X2 =X1 Y2(d)X1 = X2Y1Y2
4. Kelvin temperature can be converted into Celsius temperature by. (a) oC= K – 273 (b) k + 273 (c) 0C+273k (d) k+2730C
5. What will be the new volume (v) if the new pressure is halved and the initial pressure remains the same? (a) 2p1 V1 = p2 V2 (b) p1 V1 =2p2 V2 (c) P1 V12 = P2 V22 (d) p1 V1 =P2 V22

 

ESSAY QUESTIONS

1. 130cm3 of a gas at 200C exert a pressure of 750mmHg. Calculate its volume, is increases to 150cm3 at 350C
2. Draw the graphical representation of both Boyle’s and Charles’ law.
3. Convert the following temperature to K. (a) 150C  (b)2750C   (c) 880C
4. The volume of gas at 250C (298k) is 100cm3. What will be the volume at (a.) 750C (348k) (b). 500C (223)k, pressure remaining constant?

 

WEEKEND ASSIGNMENT: 

Read about Graham’s law, Avogadro’s number and the mole concept.