Module Heat

7/28/2019 Module Heat

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CHAPTER 4: HEAT4.1 UNDERSTANDING THERMAL EQUILIBRIUM

Heat energy has long been used by human beings for various purposes such as cooking, lighting, andkeeping warm. Heat energy is also used on a big scale in industries. A good understanding of heat will help

us manage heat energy in ways that are efficient and beneficial.

4.1 THERMAL EQUILIBRIUM

1. When two objects with different degrees of hotness come into thermal contact, heat energy istransferred between the two objects.

2. Mechanism of thermal equilibrium;

(a) Energy is transferred from the ___________ object to the ____________ object.(b) This causes the hot object A to cool down while the cold object B warms up. The temperature of

A drops while the temperature of B rises.


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(c) After some time, the rates of energy transfer between the two objects become equal. There is no

net heat transfer between the objects. The two objects are said to be in ____________________

3. Thermal equilibrium can be summarised as:(a) ______________________________________________________________________________

________________________________________________________(b) ___________________________________________________________________

4. Thermal equilibrium does not depend on the;(a) _______________________________(b) _______________________________(c) _______________________________(d) _______________________________

LIQUID-IN-GLASS THERMOMETER

1. Temperature is the measure of the ________________________ of an object.2. The SI unit of temperature is Kelvin (K).3. The temperature can be measured by a liquid-in-glass thermometer.4. Mercury and coloured alcohol are the usual thermometer liquids.5. Mercury is used in a thermometer because it;

(a) ____________________________

(b)

____________________________(c) ____________________________(d) ____________________________(e) ____________________________

6. Liquid-in-glass thermometer uses the Celsius scale, 0C. The Celsius scale is defined by two fixed pointsof temperature, which are the ___________ and the ___________

7. The lower point is defined the ___________________________________ under atmospheric pressure.

8. The lower point is the ___________________________________ under standard atmospheric pressure.


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9. A new thermometer without a scale marked on the stem can be calibrated by placing the bulb in ice and

then in steam. The corresponding levels of the mercury column are marked. The distance between theice point (0 0C) and the steam point (100 0C) is then divided into 100 equal divisions. The thermometercan now be used to measure temperature.

10. The calibrating of the thermometer can be carried out by using the following formula:

Temperature, =0100

0

_lll_l

x 100


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Example:

Given an alcohol in glass thermometer which has not been calibrated. The length of the alcohol when it isinserted into melting ice and boiling water are 4.2 cm and 12.7 cm respectively. When it is inserted into acup of warm water, the alcohol length is 8.7cm.What is the temperature of the warm water?

Solution:

Temperature, =0100

0

l_ll_l x 100

= 8.7 4.2 x 10012.7- 4.2

= 52.9 0 C


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4.2 UNDERSTANDING SPECIFIC HEAT CAPACITY

A pail of water from a swimming pool has been left beside the pool by some children in the morning. Atnoon, a lifeguard puts one hand into the water of the swimming pool and the other hand into the pail of water. He observes that the water temperatures are different.

In another situation, when equal masses of water and cooking oil are heated with similar flame, thetemperature of one will increase faster than the other.The water in the swimming pool and in the pail are said to have different heat capacities . In the lattersituation, the water and the cooking oil have different specific heat capacities .

SPECIFIC HEAT CAPACITY

1. The heat capacity of an object is the _____________________________________________________________________________________________________________

Figure 1: Three objects with different heat capacities.

2. Figure 1 shows three objects with different heat capacities because different amounts of heat arerequired to raise their temperature by 1 0C. Note that although A and B are both made of aluminium.They have different heat capacities because the masses are different.


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Figure 2: Different materials have different specific heat capacities

Figure 2 shows the amounts of heat required to raise the temperature by 1 0C for equal masses of theobjects. Note that although B and C have equal masses, different amounts of heat are required to raisethe temperature by 1 0C because they are made of different materials. Different are said to havedifferent specific heat capacities.

3. The heat capacity of an object depends on the;(a) _________________________________(b) _________________________________

4. The specific heat capacity of a substance is the ____________________________________________________________________________________________________________________________________________________________________________

5. The quantity of heat absorbed or lost by a body is given by:

6. Specific heat capacity, c can be calculated from the amount of heat supplied, Q to a mass, m of asubstance and the increase in the temperature, ,

Specific heat capacity, c =m

Q

The SI unit for specific heat capacity is J kg -1 0C -1.

7. For example the specific heat capacity of water is 4200 J kg -1 0C -1. This means that______________________________________________________________________

8. Some substances have low specific heat capacities while some have high specific heat capacities.

Substance Specific heat capacity

Aluminium 900Copper 390

Iron 470Glass 670

Paraffin 2100

Example 1The bulb of thermometer contains 3.4g of mercury. What amount of heat is required to raise thetemperature of mercury from 30 0 C to 100 0 C?


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Solution

Mass of mercury, m = 3.4 x 10 3 kg, specific heat capacity, c = 139Jkg 1 0 C 1 , change of temperature, = 100 30 =70 0 C. Amount of heat required,Q = mc

= 3.4 x 10 3 kg x 139Jkg 1 0 C 1 x 70 0 C.= 33J

Example 2320 g of hot water at 80 0 C is poured into an aluminium can of mass 35g and temperature of 20 0 C.Calculate the final temperature.

Solution Let the final temperature be T Assuming no heat loss to the surroundings, Heat lost by water = heat gained by the aluminium can

m 1 c 1 = m 2 c 2 0.320 x 4200 x (80 T) = 0.035 x 900 x (T 20)

T = 78.630

C.


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Example 1Calculate the total heat that is observed by a copper block of mass 500 g and which has been heatedfrom 31 0C to 80 0C

(Specific heat capacity of copper = 390 J kg-1

0

C-1

)

Example 2A metal of mass 2000 g is heated to 85 0C and placed in 500 g of water which is at temperature of 300C. When a steady state is achieved, the temperature of the mixture is 40 0C. Find the specific heatcapacity of the metal. (specific heat capacity of water is 4200 J kg -1 0C -1)

Example 3An electric kettle of power 2500 W contains of mass 3 kg and which is at a temperature of 25 0C.What is the time required to heat the water until it reaches its boiling limit at 100 0C.


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Example 4300 g of water at temperature 40 0C is mixed with 900 g of water at temperature 80 0C. If there is noheat loss to the surroundings, what is the final temperature when thermal equilibrium is achieved bythe mixture of water?

Example 5An electric heater is installed in the bathroom of Ramlis house. It can increase the temperature of the tap water from 15 0C to 45 0C.The heater can provide 3 kg of hot water every minute.Find the energy supplied to the water every minute. (specific heat capacity of water = 4200 J kg -1 0C -1)

APPLICATION OF SPECIFIC HEAT CAPACITY

Cooking Pot

1. The body, base and handle of the cooking pot in figure above are made of materials with differentspecific heat capacities.

Part of a cooking pot Characteristic

Cooper base

Wooden handle

Aluminium body


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Claypot

1. Clay has a ______________________________________________ than metals.2. It is also a poor conductor of heat.3. During cooking, heat is conducted slowly from the fire to the food inside the pot.4. The longer cooking time brings out the taste in the food. After the flame is switched off, the claypot is

at a higher temperature than the food inside it.5. A considerable amount of heat continues to be transferred into the food. The food remains hot for a

longer period of time.

Cooling System of a Car Engine

1. Water has a large specific heat capacity. It can absorb a large amount of heat without a high increase intemperature.

2. Water is used to cool down internal combustion engines such as the car engine.


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Sea breeze and Land breeze


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4.3 Understanding Specific Latent Heat

Change of Phase

(a) (b)

1. Water exists in three phases: solid (ice), liquid and gaseous phase (steam). Photograph shows a piece of ice that has just been taken out from the freezer. The ice quickly absorbs heat from the warmersurroundings and melts.

2. The heat supplied to a substance during change of phase ________________________________________________________________ When a solid melts or a liquid isboiling, _______________ but the temperature remains _____________

Figure 1


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3. The four main changes of phase: melting, boiling, condensation and solidification are summarized infigure 1.

Figure 2: The heating curve

4. Figure 2 shows the heating curve when a substance in a solid state is heated uniformly and undergoes achange of phase from solid to liquid to gas.- A to B : ______________

According to the kinetic theory of matter, molecules ina solid are arranged in regular patterns and are held together tightly. When thesolid is heated, the molecules _________________ and _______________ abouttheir fixed positions at greater speeds and greater amplitudes. The___________________ of the molecules increases and the ____________ of the

solid ______________

- B to C : ______________

The solid begin to _____. ______________________ is ______________ by thesolid to provide the energy to___________________________________________. The molecules are freedfrom their fixed positions and are able to move about. The energy absorbed doesnot increase the _______________ of the molecules, so the temperature____________________ during melting.

- C to D : ______________

The liquid __________________. The heat supplied__________________________ of the liquid molecules which begin to movefaster and randomly.


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- D to E : ______________

The liquid starts to ______. ____________________ is absorbed by the liquid.This provides the liquid with the _____________________________________between the molecules and to work against the atmospheric pressure as the liquidexpands into gas.

- E to F : ______________

The liquid vaporized completely into a gas. The heat supplied increase the kineticenergy of the gas molecules and its temperature rises.

Figure 3: The cooling curve

- A to B : _____________When the gas is cooled, ________________ to the surroundings and the kineticenergy of the molecules ___________, resulting in a fall in the temperature of the gas.

- B to C : ______________

The temperature__________________ during this change in state and the latentheat of evaporations is ______________ to the surroundings.

- C to D : _____________

The average kinetic energy of the liquid molecules decreases resulting in a fall inthe temperature of the liquid until the solidification points.


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- D to E : The molecules become more closely packed as the liquid solidifies. Thetemperature remains constant during this change in state and the latent heat of fusion is released to the surroundings.

- E : SolidOn further cooling, the average kinetic energy of solid molecules decreasesresulting in a fall in the temperature of the solid.

Latent Heat

1. The ____________ or the ________________ at ______________ _____________ during a_______________ is known as latent heat.

2. This transfer of heat _____________________________________ of the molecule.

Latent Heat of Fusion

Figure 3

1. During melting, the heat absorbed is used to __________ the bonds between the particles.

2. The particles are freed from their fixed positions and are able to vibrate and move among each other asillustrated in figure 3.

3. The ____________________ is known as the latent heat of fusion . For a liquid to solidify at its

freezing point, latent heat of fusion has to be removed.


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Latent Heat of Vaporisation

Figure 4

1. When a liquid boils, the heat absorbed is used to completely break the bonds between the particles.

2. The __________________________ is known as the latent heat of vaporisation .

3. When vapour condenses back into the liquid phase, latent heat of vaporisation is released.

Specific Latent Heat

1. The specific latent heat of a substance is the ______________________________________________________________________________________________________________________

The specific latent heat,

Q = latent heat absorbed or released by the substance.m = mass of the substanceWith unit in J kg -1

2. The specific latent heat of fusion of a substance is defined as the______________________________________________________________________________________________________________________________________________________

3. For example the specific latent heat of fusion of ice is 336 000 J kg -1. This mean ____________________________________________________________________________________________

4. The specific latent heat of vaporisation of a substance is defined as the______________________________________________________________________________________________________________________________________________________


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Example 1An electric kettle contains 3.4 kg of water at 30 0C.(a) Calculate the amount of heat required to boil away all the water after the boiling point has been

reached.(b) If the power of the heater is 2.4 kW, what is the time taken?[Specific heat capacity, c = 4200 Jkg -10C-1, Specific latent heat of vaporisation of water = 2.26 x 10 6 J kg -1]

Example 2Water of mass 0.36 kg at 25 0C is put into the freezer compartment of a refrigerator. Calculate the amount of heat that must be removed to change the water completely into ice.[Specific heat capacity, c = 4200 Jkg -10C-1, specific latent heat of fusion = 3.36 x 10 5 J kg -1]

Example 3What is the quantity of heat that is required to convert 40g of ice into steam at 100 0C.[Specific heat capacity, c = 4200 Jkg -10C-1, specific latent heat of fusion = 3.36 x 10 5 J kg -1, Specific latentheat of vaporisation of water = 2.26 x 10 6 J kg -1]


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Experiment: Determine the specific latent heat of fusion of an ice.

Method:1. Two sets of equipment are arranged as in figure.2. The first set off apparatus is supplied with electrical power, whereas the second set of the apparatus is a

control set without electrical power supply.3. The mass of each of the two empty beakers, A and B is determined using the weighing balance.4. A stop watch is started simultaneously as the current supply is turned on.5. After a suitable time interval, the electrical current is turned off.6. The masses of water in both beakers are determined.

The control set of apparatus was set up so that the mass of ice melted by the heat supplied by theheater only can be determined.The value of the specific latent of fusion of ice obtained is larger than the standard value. Some of the heat supplied by the heater is lost to the surroundings.


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Exercises

1. What is the quantity of heat required to change 0.7 kg of water at 27 0 C to steam at 100 0C?[Specific heat capacity of water, c = 4200 Jkg -10C -1, Specific latent heat of vaporization of water, l =2.26 x 10 6 Jkg -1]

2. A 500W electric heater is used to boil water. What is the time required to reduce the mass of waterby 80 g after the water has reached its boiling point?[Specific latent heat of vaporization of water, l = 2.26 x 10 6 Jkg -1]

3. By using an immersion heater marked 240 V, 3 kW in an experiment, calculate the time taken toconvert 3 kg of boiling water into steam.[Specific latent heat of vaporization of water, l = 2.26 x 10 6 Jkg -1]

4. A piece of ice of mass 2 kg at a temperature of 0 0C is heated until it becomes water at 10 0C. If thespecific heat capacity of water is 4200 Jkg -10C-1 and the specific latent heat of fusion of ice is 3.36 x105 Jkg -1, what is the energy supplied?

5. In an experiment, Azmi uses an electric heater to heat 1.0 kg of ice at 0 0C until it becomes water at600C. Calculate the electrical energy used in the experiment.[Specific heat capacity of water = 4.2 x 10 3 Jkg -10C -1, specific latent heat of fusion of ice = 3.34 x 10 5 J kg -1]

6. A piece3 of ice is added to 200 g of water at 30 0C. When all the ice has melted, the temperature of the water becomes 20 0C. What is the mass of the ice?[Specific heat capacity of water = 4.2 x 10 3 Jkg -10C -1, specific latent heat of fusion of ice = 3.34 x 10 5 J kg -1]


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7. On a hot day, a student added 80 g of ice into 250 g of orange squash at a temperature of 28 0C. Whatis the final temperature of the orange squash?[Specific heat capacity of the orange squash = 4.2 x 10 3 Jkg -10C -1, specific latent heat of fusion of ice= 3.34 x 10 5 J kg -1]

8. 1.2 kg of water at a temperature of 28 0C is put into the freezer of a refrigerator of 1200 W. How longdoes it take for all the water to completely freeze to ice?[Specific heat capacity of water = 4.2 x 10 3 Jkg -10C -1, specific latent heat of fusion of ice = 3.34 x 10 5 J kg -1]


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4.4 UNDERSTANDING THE GAS LAWS

The idea of Gas Laws

Figure 1.1 Figure 1.2

You feel a resistance when you try to compress aballoon.

The air in the tyre of a car is able to support theweight of the car.

Figure 1.3 Figure 1.4

The lid of an empty tin bursts open when the tinis dropped into a fire.

The plastic bottle bulges after it has beenexposed to strong sunlight.

1. The observation of the events in figure above show that a gas has volume and temperature and it exertsa pressure.

2. These properties are explained in terms of the behaviour of gas molecules in table 1.

Property of a gas Explanation based on the kinetic theory

Volume

Temperature

Pressure

Table 1


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3. Table 2 gives a summary of the units for pressure, volume and temperature of a gas.

Property and symbol SI unit Symbol for SI unit Other units

Pressure, P N cm -2, cm Hg

Volume, V mm3

, cm3

Temperature, T 0C

Table 2

Relationship between pressure and volume

Figure 2

1. When the volume of a gas is decreased, the number of molecules______________________________________________________________________________________________________________________

2. The same number of molecules moves in a smaller space.

3. The molecules collide more frequently with the walls of the container.

4. The _____________________________________________ results in an _____________ in the________________ exerted by the gas.


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BOYLES LAW

1. Boyles law states that for a fixed mass of gas, the _____________ of the gas is_______________________ to its _________ when the temperature and mass is kept constant.

In smaller volume, molecules hit the walls more often resulting in greater pressure.

2. The relationship between pressure and volume can be explained using the kinetic theory of gases :

(a) When the volume of a gas is decreased, the number of molecules per unit volume increases.

(b) The same number of molecules moves in a smaller space.

(c) The molecules collide more frequently with the walls of the container.

(d) The increases in the rate of collision results in an increase in the pressure exerted by the gas.

Experiment: To determine the relationship between pressure and volume of a fixed mass of gas at constanttemperature.

Variables:

Manipulated variable: ____________

Responding variable: _____________

Fixed variables: Mass of the air in the tube and the temperature

Apparatus:


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3. The mathematical expression for Boyles law is,

4. Boyles law can be shown graphically as in figure 2

The relationship between pressure and volume can also be expressed

with the graphs above.

5. An example of Boyles law is the action of the diaphragm of our body. When we inhale, the diaphragmmoves downward and increases the volume of the lungs so that the pressure is less than the atmosphericpressure. This results in air being forced into the lungs. When we exhale, the diaphragm moves upwardand decreases the volume of the lungs. This increases the pressure inside the lungs above theatmospheric pressure on the outside so that gases are forced out of the lungs.

CHARLES LAW

1. Charles law states that for a fixed mass of gas, the ________ of the gas is ______________ to itsabsolute _________ when its pressure is kept constant.

2. The relationship between volume and temperature can be explained using the kinetic theory of gases.

(a) When gas is heated, the average __________ of the molecules ___________. The ___________ of the gas increases.


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(b) The _______________ between the molecules and the _______ will _______ if the volume isconstant.

(c) If the gas is allowed to expand, the faster molecules now move in a bigger space,

(d) Therefore, the rate of collision between the molecules and the walls remain constant and thus thepressure is constant.

Experiment : To determine the relationship between volume and temperature of a fixed mass of gas atconstant pressure.

Variables:

Manipulated: ______________

Responding: _______________

Fixed: Mass of the air in the tube and the pressure.

Apparatus:

3. The mathematical expression for Charles law is,


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4. The volume of gas decreases when its temperature is lowered. At 0 0C, the gas has a certain volume. It isimpossible to lower the temperature of the gas until the volume becomes zero.

Figure 3

5. By extrapolating the graph of volume against temperature. It is found that at -273 0 C the volume isexpected to become zero, as shown in figure 3.

6. The temperature -2730

C is the lowest possible temperature and in known as the______________________________.

7. SI unit for temperature is Kelvin (K). Temperature measured in the kelvin scale are known as absolutetemperatures. Absolute zero is given the value 0 kelvin (0 K). Thus, in the Kelvin scale, there are nonegative values of temperature.

8. When the Kelvin scale is used for temperature, the graph of volume against absolute temperature is astraight line passing through the origin.

9. The corresponding temperatures in the Celsius and Kelvin scale are as follows:

Temperature Celsius scale Kelvin scale

Absolute zero -273 0 C O K

Ice point 0 0 C 273 K

Steam point 100 0 C 373 K

Unknown temperature 0 C ( + 273) K


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PRESSURE LAW1. The pressure law states that for a fixed mass of gas, the _________ of the gas is __________________

to its ____________________ when its volume is kept constant.

2. The relationship between pressure and temperature can be explained using the kinetic theory of gases:

(a) When a gas is heated, the average _________________ of the molecules __________________.The _____________ of the gas increases.

(b) The faster moving molecules strike the walls of the container more frequently.

(c) The molecules also experience a larger ________________________ when they bounce back from the walls.

(d) A larger force is exerted on the walls resulting in a higher pressure.

Experiment: To determine the relationship between pressure and temperature of a fixed mass of gas atconstant volume.

Variables:

Manipulated: ______________

Responding: ______________

Fixed: The mass of the air in flask and the volume

Apparatus:


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3. When the temperature of the gas is reduced to 0 0C, the molecules still have kinetic energy which is lessthan that at room temperature. The gas exerts a lower pressure.

4. The temperature at which the pressure of the gas is expected to becomes zero can be obtained byextrapolating the graph of pressure against temperature. It is found that at -273 0C, the pressure becomeszero.

5. The graph of pressure against absolute temperature is shown in figure above.

6. The mathematical expression for the Pressure law is,


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Problems Solving of Gas LawsBoyles Law

1. An air bubble of capacity 15 cm 3 at a pressure of 320 cm Hg is released from the sea. Determine thevolume of the bubble when it reaches the surface of the water, if the atmospheric pressure is 76 cmHg. Assume that the temperature of the water is constant.

Solution Initial volume, V 1 = 15 cm 3 Initial pressure, P 1 = 320 cm HgFinal volume, V 2 = ?Final pressure, P 2 = 76 cm Hg

P 1V 1 = P 2V 2 V 2 = P 1V 1 /P 2

= (320)(15)/(76)= 63 cm 3

2.

A bicycle pump of length 40 cm has a uniform cross-sectional area. It contains air at atmosphericpressure. If the piston is pushed down a distance of 17 cm to force air into a tyre, what is theresultant air pressure in the pump? Assume that the temperature of the air is constant. [P atm = 1 atm]

SolutionP 1V 1 = P 2V 2 P 2 = (1)(40)/(23)

= 1.74 atm


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3. Figure 4.4.2 shows a glass tube closed at one end with a small quantity of mercury trapped inside.When the tube is in a horizontal position, the length of air column at the closed end of the tube is 8cm and the length of the mercury column is 2 cm.

(e) What is the pressure of the air trapped inside the tube in a horizontal position?(f) What is the length of the column of air trapped inside the tube, when the tube is in a vertical

position as shown in figure 4.42(b)?

Solution(a) When the glass tube is in a horizontal position, the pressure of the air trapped inside is the

same as the pressure of the air outside, i.e. atm pressure.Pressure of the trapped air, P = 76 cm Hg.

(b) P 1V 1 = P 2V 2 V 2 = [P 1V 1]/P 2

= [76 x 8]/78 = 7.8 cm

i. A rod of tube with one end closed has a strip of mercury 3 cm long. When the tube is held vertically,the length of air trapped is 8 cm as in figure (a) above. Atmospheric pressure is taken as 76 cm Hg.(a) Calculate the length, x of air which is trapped when the tube is placed horizontally as in

figure (b) above.(b) Calculate the length, y of air which is trapped when the tube is inverted as shown in figure (c)

above.


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Solution(a) P 1V 1 = P 2V 2

(76 + 3)8 = 76 (x)x = 632/76

= 8.3 cm

(b) P 1V 1 = P 2V 2 (76 + 3) 8 = (76 3) y

y = 632/73= 8.7 cm

ii. A bubble of air with volume 4 cm 3 is released from a submarine at the depth of 50 m. What will bethe volume of the bubble when it has risen to a depth of 14 m?[Atmospheric pressure = 10 m water]

SolutionP 1V 1 = P 2V 2 60 x 4 = 24 x V 2

V 2 = (60 x 4)/24= 10 cm 3

iii. An air bubble released by a diver has a volume of 6.0 cm 3 at a depth of 5 m. What is the volume of the bubble at a depth of 2 m?[Assume atmospheric pressure is equivalent to 10 m of water]

SolutionP 1V 1 = P 2V 2 (5 + 10) x 6 = (2 + 10) V 2

V 2 = [15 x 6]/12

V 2 = 7.5 cm3


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Figure 1

iv. Figure 1 shows trapped air inside a thin glass tube in three different positions. Determine the value of x and y.[Atmospheric pressure = 76 cm Hg]

SolutionPosition (a)P1 = (76 + 4) cm Hg

= 80 cm Hg

Position (b)P2 = (76 4) cm Hg

= 72 cm Hg

Position (c)P3 = 76 cm HgComparing (a) and (b)P 1V 1 = P 2V 2 80 x 12 = 72 x x

x = (80 x 12)/72 x = 13.3 cm

Comparing (a) and (c)P 1V 1 = P 2V 2 80 x 12 = 76 x yy = (80 x 12)/76

= 12.6 cm


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Pressure Law

v. An iron cylinder containing gas has a pressure of 360 kPa when it is kept in a store at temperature270C. What is the pressure of the gas when the cylinder is moved outdoors where the temperature is400C.

SolutionP 1 = 360 kPa T 1 = (27 + 273) = 300 K P 2 = Final pressure T 2 = (40 + 273) = 313 K

Using the pressure law,P 1 /T 1 = P 2 /T 2 360/300 = P 2 /313P 2 = (360 x 313)/300P 2 = 375.6 kPa

vi. Before a journey from Alor Setar to Ipoh, the air in a car tyre has a pressure of 200 kPa and atemperature of 27 0C. After the journey, the air pressure in the tyre is 240 kPa. What is the

temperature of the air in the tyre after the journey?[Assume the volume of a tyre is constant]

SolutionP 1 /T 1 = P 2 /T 2 200/(27 + 273) = 240/T 2 T 2 = (300 x 240)/200T 2 = 360 K T 2 = 360 273

= 87 0C

vii. The pressure of nitrogen gas in a light bulb is 60 kPa at 200

C. Calculate the temperature of the gaswhen the pressure inside the bulb rises to 90 kPa after the bulb is lighted up.

SolutionP 1 /T 1 = P 2 /T 2 60/293 = 90/T 2 T 2 = (90 x 293)/60T 2 = 439.5 K

Final temp. = 439.5 273= 166.5 0C

viii. A closed flask contains gas at a temperature of 95 0C and pressure of 152 kPa. If the temperature isreduced to -17 0C, what is the new pressure of the gas?


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SolutionP1 /T 1 = P 2 /T 2 P2 = (152)(256)/368P2 = 106 kPa

ix. A motorcycle tyre is found to posses a pressure of 3.2 atmospheric when it was tested in a garage at atemperature of 27 0C. By assuming that the volume of the air in the tyre is fixed, what is thesurrounding temperature if the tyre is fixed, what is the surrounding temperature if the tyre is foundto posses a pressure of 2.8 atmospheric?

SolutionP 1 /T 1 = P 2 /T 2 3.2/300 = 3.5/T 2 T 2 = 328 K

The surrounding temp. = (328 273)= 55 0C

Charles Law

x. A syringe in a refrigerator contains 4.5 ml of air at -3 0C. When the syringe was taken out and placedin a room where the temperature was 27 0C, the air in it expands. Calculate the final volume of the airin the syringe.

SolutionV 1 /T 1 = V 2 /T 2 4.5/270 = V 2 /300V 2 = (4.5 x 300)/270V 2 = 5.0 ml

xi. The volume of a fixed mass of gas is 150 cm 3 at 35 0C. If the gas is heated at a constant pressure,what is the volume of the gas when the temperature reaches 87 0C?

xii. The figure above shows a glass tube containing some trapped air inside it. At 27 0C, thehorizontal column of trapped air is 9 cm.If the atmospheric pressure does not vary, how much is the horizontal column of air at a temperatureof 45 0C?

SolutionV1 /T 1 = V 2 /T 2 9A/300 = xA/318

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