Chapter Ten
GAS LAWS
Boyle’s Law
It states that the pressure of a fixed mass of a gas is inversely proportional to its volume, provided the temperature is kept constant.
Stated in symbols;
P α ^{1}/_{V} OR P= k x ^{1}/_{V}
So, PV = constant
How to proof the Boyles law
Apparatus
Thick-walled J-shaped glass tube with one end closed, oil, Bourdon gauge, foot pump, meter rule.
Procedure
- Set up the apparatus as shown above
- Connect the foot-pump to the apparatus and with the tap open, pump in air until the oil rises a small but measurable height, then close the tap.
- Allow the air to adjust to room temperature, then read the value of pressure on the gauge and the height h of air column, which represents volume of the air (glass tube has uniform cross-sectional area).
- Repeat the experiment by varying values of pressure to obtain at least five more readings.
Record your results in table
Table 10.1
Pressure p(Pa) | Volume h (cm) | ^{1}/_{v} (^{1}/_{h})cm-1 | pv |
Pressure, p(x 10^{5}Pa) | 2.00 | 2.50 | 3.00 | 3.50 | 4.00 | 4.50 |
Volume, V (m^{3}) | 0.025 | 020 | 0.017 | 0.014 | 0.012 | 0.011 |
- Using your results in table plot a graph of:
(i) P against V.
(ii) P against ^{1}/_{v}. Determine the slope.
(iii) PV against P.
Results and Conclusion
The experiment show that an increase in pressure of a fixed mass of a gas causes a decrease in its volume.
The sketches below show the relationship between pressure P and volume V of a fixed mass of gas.
The graph of P against V is a smooth curve while that of P against1/v is a straight line passing through the origin. That of PV against P is a straight line parallel to the x-axis. Since PV = constant;
P I V I = P 2 V 2 = constant, for any given mass of a gas.
Example 1
The diagram below shows an air bubble of volume 2.0 cm^{3} at the bottom of a lake 40 m deep.
What will be its volume just below the surface S if the atmospheric pressure is equivalent to a height of 10 m of water?
Solution
10m height = 1 atm.
40 m height = 4 atm.
Pressure PI at the bottom = (l + 4) =5 atm
Pressure P, at surface = 1 atm
Volume V: at bottom = 2 cm^{3}
By Boyle’s law, PIV_{I} = P_{2}V_{2}
5 x 2 = 1 X V_{2}
V_{2}= 10
Volume just below surface is 10 cm^{3},
Example 2
The volume V of a gas at pressure P is reduced to ^{3}/_{8}V without change of temperature. Determine the new pressure of the gas.
Solution
PV = constant
P1V1, =P_{2}V_{2}
P1V1 = P_{2} x ^{3}/_{8}V_{1 }
P_{2} =^{8}/_{3}P_{1}
Example 3
A column of air 26 cm long is trapped by mercury thread 5 cm long as shown in (a). When the tube is inverted as in figure (b), the air column becomes 30 cm long. What is the value of atmospheric pressure?
Solution
In (a), gas pressure = atm pressure + hƥg,
In (b), gas pressure = atm. pressure – hƥg,
where ƥ is the density of mercury.
From Boyle’s law;
P_{1} V_{1} = P_{2}V_{2 }
Let the atmospheric pressure be height ‘x’ of mercury.
So, (x + 5) 0.26 == (x – 5) 0.30
0.26x + 1.30 == 0.3x – 1.5
2.8 = O.04x
:. x = 2.8
0.04
=70cm
If experiment above is repeated at different temperatures, similar curves are obtained as in figure each is called an isothermal curve.
When P is plotted against ~ for each of the isothermals, the figure below is obtained.
Sample questions
P |
V |
- On the axes provided sketch the P-V graph for a gas obeying Boyle’s law.
- 2003 Q24 P1
- On the axes provided in, sketch a graph of pressure(p) against reciprocal of volume(^{1}/_{V}) for a fixed mass of an ideal gas at constant temperature.
p
^{1}/_{V}
- The pressure acting on a gas in a container was changed steadily while the temperature of the gas maintained constant. The value of volume V of the gas was measured for various values of pressure. The graph in figure 9 shows the relation between the pressure, p and the reciprocal of volume 1
v
- Suggest how the temperature of the gas could be kept constant
- Given that the relation between the pressure P_{1} and the volume, V_{1} of the gas is given by PV = k When k is a constant, use the graph to determine the value of k.
- What physical quantity does k represent? (4 marks)
- State one precaution you would take when performing such an experiment (1 mark)
(c) A gas occupies a volume of 4000 litres at a temperature of 37^{0}C and normal atmospheric pressure. Determine the new volume of the gas if it heated at constant pressure to a temperature of 67^{0}C (normal atmospheric pressure P= 1.01 x 10^{5} pa)
- An air bubble is released at the bottom of a tall jar containing a liquid. The height of the liquid column is 80cm. The volume of the bubble increases from 5 cm^{3} at the bottom of the liquid to 1.15 cm^{3} at the top. Figure below shows the variation of pressure, P, on the bubble with the reciprocal of volume 1/v^{3} as it rises in the liquid.
(i) State the reason why the volume increases as the bubble rises in the liquid column (1 mark
(ii) From the graph, determine the pressure on the bubble:
(I) at the bottom of liquid column; (2 marks)
(II) at the top of the liquid column (1 mark )
(iii) Hence determine the density of the liquid in kgm^{3} (3 marks)
(iv) What is the value of the atmospheric pressure of the surrounding? (1 mark)
(c) A rubber tube is inflated to pressure of 2.7 x 10^{5} Pa and volume 3800 cm^{3 }at a temperature of 25^{◦}C. It is then taken to another place where the temperature is 15⁰C and the pressure 2.5 x 10^{5} pa. Determine the new volume. (4 Marks)
Charles’ Law
It states that the volume of a fixed mass of gas is directly proportional to its absolute temperature if the pressure is kept constant.
The set-up in figure can be used to demonstrate the relationship between temperature
and volume of a given mass of a gas at constant pressure.
The flask is grasped firmly and the water index observed.
The water index rises higher when the flask is held and falls when the hands are withdrawn, showing that the volume of gas increases when its temperature is raised.
EXPERIMENT: To investigate the relationship between volume and temperature of a given mass of gas at constant pressure
Apparatus
Capillary tube sealed at one end, cone, sulphuric acid, thermometer, rule, and stirrer, source of heat, retort stand, rubber band, and water bath.
Procedure
Introduce concentrated sulphuric acid deep into the glass tube to trap air in the tube.
Attach the tube, thermometer and the ruler using rubber band.
Assemble the apparatus as shown in figure above.
Record the room temperature and the corresponding height h of air in the tube.
Heat the bath and record the temperature and the height at suitable temperature intervals in table below.
Temp (˚c) | ||||
Height h(cm) |
Note:
(i) The sulphuric acid index serves as a pointer to the volume of the gas on the scale as well as a drying agent for the air.
(ii) Pressure of the trapped air is the same as the atmospheric pressure plus pressure due to the acid index. This remains constant throughout the experiment
(iii) Before taking the readings. Stir the bath so that the temperature of the gas is equal to that of the bath.
- Plot a graph of volume (height h cm) against temperature.
Observations
As the temperature rises, the height h (volume) also increases.
The graph is a straight line, indicating proportional changes in volume and temperature.
- the graph does not pass through the origin.
- If the graph is extrapolated it cuts the temperature axis at about -273⁰ At this temperature the volume of the gas is assumed to be zero.
- This temperature -273 ˚C. at which the volume of a gas is assumed to be zero is the lowest temperature a gas can fall to. It is therefore called absolute
The scale of temperature based on the absolute zero is called the absolute scale or Kelvin scale of temperature.
A plot of volume against absolute temperature gives a straight line through the origin, as shown in figure
Note:
It is impossible to get to absolute zero for gases because they condense at fairly higher temperatures.
In symbols, Charles ‘law can be stated as follows;
V αT or V = kT, where k is a constant.
Hence, ^{V}_{1}/_{T1} = ^{V2}/_{T2}constant
Note:
This formula is only applicable when T is expressed in Kelvin.
Relation Between Celsius and Absolute Scale
Figure 10.12 relates the absolute (Kelvin) scale to the Celsius scale.
The zero Kelvin (0 K) corresponds to -273°C while 0˚C corresponds to 273 K.
It follows that to change from Celsius to Kelvin, we add 273 to the Celsius temperature, i.e.;
= (T + 273) K
Example 3
The temperature of a gas is -42 ⁰C. What is this temperature on the Kelvin scale?
Solution
Example 4
0.02 m^{3} of a gas af27 °C is heated at constant pressure until the volume is 0.03 m^{3}. Calculate the final temperature of the gas in “C.
Solution
Example5
A mass of air of volume is 750 cm^{3} is heated at constant pressure from 10 °C to 100°C. What is the final volume of the air?
Solution
a)What is meant by absolute zero temperature? (1 mark)
- b) 10 shows a set up to investigate the relationship between temperature and volume for a certain gas.
- b) State two factors that are kept constant, in order to determine the relationship. (2 marks)
- c) The graph in Fig. 11 shows the relationship between volume and temperature for the experiment.
(i)What was the volume of the gas at 0^{0}C? (1 mark)
(ii)At what temperature would the volume of the gas be zero? (1 mark)
(iii) Explain why the temperature in part (ii) above cannot be achieved. (2 marks)
- d) A sealed gas cylinder contains 300cm^{3} of certain gas at a temperature of 25^{0}C, and at a pressure of 9.5x 10^{4}pa the gas in the cylinder was then cooled to 10^{0}
Determine the new pressure of the gas in the cylinder. (4 marks)
- 2009 Q13 P1
State what is meant by absolute zero temperature (Zero Kelvin or -273^{◦}C). (1 mark)
Pressure law
It states that the pressure of a fixed mass of gas is directly proportional to its absolute temperature, provided the volume is kept constant.
EXPERIMENT: To investigate the relationship between the pressure and the temperature of a fixed mass of gas at constant volume
Apparatus
Round-bottomed flask, pressure gauge, water bath, heater, thermometer.
Procedure
- Set up the apparatus as shown in figure above.
- Record the initial temperature and pressure readings.
- Heat the water bath gently and obtain at least seven more pairs of readings at suitable temperature intervals.
- Record your results in table 10.3.
Table 10.3
Temp (c | |||||||
Pressure(pa) | |||||||
Pressure
Temp |
Note:
The air in the tube connecting the gauge to the flask may be at a lower temperature than the air in the flask. This tube should therefore be as short as possible.
Observations
It is observed that increase in temperature causes an increase in pressure.
A plot of pressure against temperature gives the graph shown in figure.
Graph of pressure against temperature
When the graph is extrapolated, it cuts the temperature axis at -273°C, the absolute zero.
Figure shows the same graph of an absolute temperature scale.
On the absolute scale, the pressure of a gas is directly proportional to its absolute temperature.
This conclusion is summed in pressure law which
In symbols;
PαT (V constant)
Or P = kT, where k is constant
So P_{1} = P_{2}
, T_{1} T2
Example 6
A cylinder contains oxygen at 0⁰C, and 1 atmosphere pressure. What will be the pressure in
the cylinder if the temperature rises to 100 °C?
Solution
= 1.37 atmosphere
Example 7
At 20°C, the pressure of a gas is 50 cm of mercury. At what temperature would the pressure of the gas fall to 10cm of mercury?
Solution
= 58.6 K (or -214.4 “C)
- a) Figure 6 shows a simple set up for pressure law apparatus.
Describe how the apparatus may be used to verify pressure law. Initial reading of pressure and temperatures are recorded
Water is heated and gently stirred;
Values f pressures and temperature are recorded to intervals;
Temperature is converted to K and atmospheric pressure p added to P;
Graph of pressure p against temperature (K) Plotted giving straight line;
Equation of State
A general gas law relating the changes in pressure, volume and the absolute temperature can
be derived from the three gas laws.
Consider a fixed mass of gas which is being changed from state A to state B through an
intermediate state C, as shown in figure.
Charles’ law Boyle’s law
From A to C, the gas is heated at constant pressure PI’ By Charles’ law;
V_{1} = V_{c}
T_{1} T_{2}
Volume V_{c} in state c, V_{c} = V_{1}T_{2}
T_{1}
From C to B, the gas pressure is changed from P_{1 }to P_{2} at constant temperature T_{2}. By Boyle’s law,
P_{1}V_{c} = P_{2}V_{2}
Vc = P_{2}V_{2}
P_{1}
_{ }
Equating the two
V_{1}T_{2} = P_{2}V_{2}
T_{1} P_{1}
Rearranging the equation
V_{1} P_{1 }= P_{2}V_{2}
T_{1} T_{2}
In general, PV = k, where k is a constant.
T
This is known as the equation of state, in which k depends on the type and quantity of the gas.
The equation changes to when the amount of gas is 1mole. Constant R is same for all gases, and is called the universal gas constant.
Example 8
A mass of 1 200 cm^{3} of oxygen at 27°C and a pressure 1.2 atmosphere is compressed until its volume is 600cm^{3 }and its pressure is 3.0 atmosphere. What is the Celsius temperature of the gas after compression?
Solution
102˚C
Gas Laws and Kinetic Theory
If the volume of a vessel containing a fixed mass of gas is halved, the number of molecules per unit volume will be doubled. The number of collisions per unit time, and therefore the rate of change of momentum, will also be doubled. Consequently, halving the volume of the gas
doubles the pressure, which is the import of the Boyle’s law.
Charles’ Law and Kinetic Theory
When a gas is heated, the kinetic energy and therefore the velocity of the molecules increases.
As the temperature rises, the molecules move faster. If the volume of the container were constant,
the pressure on the walls would increase due to greater rate of change of momentum per unit time. But since Charles’ law requires that the pressure be constant, then the volume must
increase accordingly so that although the molecules are moving faster, the number of collisions at the walls of the container per unit time is reduced, since the distance between the walls is increased by increasing the volume.
Pressure Law and Kinetic Theory
In gases, pressure is as a result of bombardment of the walls of the container by the gas molecules. When the molecules of the gas bombard and rebound from the walls of the container, a change of momentum takes place. The number of bombardments per unit time constitutes a rate of change of momentum, which according to Newton’s second law of motion, constitutes a force. This force per unit area emerges as the pressure of the gas.
When a gas is heated, its molecules gain kinetic energy and move about faster. If the volume of the container is constant, the molecules will bombard the walls many more times per unit time, and with greater momenta. The total rate of change of momentum will therefore increase. The resulting force per unit area, which is the pressure, will increase.
Limitations of Gas laws
When explaining the gas laws using the kinetic theory, both the size of molecules and the intermolecular forces are assumed to be negligible.
Real gases have molecules with definite volumes and therefore the idea of zero volume or zero pressure is not real. Real gases get liquefied before zero volume is reached.
This departure from the gas laws is so particularly true at low temperatures and high pressures. A gas that would obey the gas laws completely is called ideal or perfect gas.