9 First Law of Thermodynamics

Question 1

Explain microscopic kinetic energy.

Answer 1

The microscopic kinetic energy of a body is due to the kinetic energies of its particles due to their constant random motion, which can be translational, rotational or vibrational.

Question 2

Explain how temperature is related to the microscopic kinetic energy of particles.

Answer 2

The temperature of a body is dependent on the AVERAGE microscopic kinetic energy of all its particles. For an ideal gas, its temperature is directly proportional to the average microscopic kinetic energy. (In other words, the faster the movement of a body’s particles, the higher its temperature) During heating, the microscopic kinetic energy of the body’s particles increases, the average microscopic kinetic energy of the body increases. Hence, its temperature increases.

Question 3

Explain microscopic potential energy and why it is negative for liquids and solids.

Answer 3

The microscopic kinetic energy of a body is due to the potential energies of its particles, arising from the interactions between them. For an ideal gas, its microscopic kinetic energy is defined to be zero.

Since heat is required for a liquid to change into the gaseous phase, the microscopic potential energy of liquids must be lower than that of gases. Similarly, the microscopic potential energy of solids is also negative and lower than that of liquids.

When solid changes into the liquid phase, bonds are being broken and its microscopic potential energy increases. Similarly, when a liquid changes into the gaseous phase, its microscopic potential energy increases as the bonds are being broken. Hence, during melting, the temperature of an object does not change.

Question 4

Describe what happens during a change in phase during melting.

Answer 4

During melting, the lattice structure has to break. Hence, some bonds are broken. Heat is first supplied, causing the internal energy of the body to increase. The average microscopic kinetic energy of the particles and thus, its temperature increases until its melting point is reached. Here, the particles have enough microscopic kinetic energy to vibrate so violently that the attractive forces cannot hold them together. This causes the lattice structure of the solid to collapse. At the melting point, the heat supplied to the solid, known as latent heat of fusion, is used to overcome the attractive forces between the atoms or molecules. The microscopic potential energy of the particles increases while the microscopic kinetic energy of the particles remains the same. Hence temperature remains constant. Once the change of phase is complete, the temperature continues to rise if more heat is supplied.

Note: The change of phase is not instantaneous. During transition, the two phases coexist until the transition is completed.

Question 5

Describe what happens during a change in phase during boiling.

Answer 5

During boiling, the bonds between the particles have to be completely broken. Heat is first supplied causing the internal energy of the body to increase. This causes the average microscopic kinetic energy of the particles and thus, its temperature to increase until its boiling point. Heat supplied, also known as the latent heat of vapourisation, is used to overcome the attractive forces between the molecules until the bonds are completely broken and a change of phase occurs. The microscopic potential energy of the particles increases while the microscopic kinetic energy of the particles remains the same. Hence temperature remains constant. Once the change of phase is complete, the temperature continues to rise if more heat is supplied.

Note: The change of phase is not instantaneous. During transition, the two phases coexist until the transition is completed.

Question 6

Define heat capacity. State the formula which expresses heat capacity.

Answer 6

The heat capacity of a body is the quantity of heat required to cause a unit rise in the temperature of the body. This can be expressed as C = Q/Change in T, where its SI unit is JK^(-1).

Question 7

Define specific heat capacity. State the formula which expresses specific heat capacity.

Answer 7

The specific heat capacity of a body is the quantity of heat required per unit mass to cause a unit rise in the temperature of the body. This can be expressed as c = Q/mChange in T, where its SI unit is Jkg^(-1)K^(-1).

Question 8

Define latent heat.

Answer 8

Latent heat is the amount of heat involved when a body changes phase.

Question 9

Define specific latent heat of fusion.

Answer 9

Specific latent heat of fusion is the quantity of heat required to convert the unit mass of solid to liquid without any change in temperature.

Question 10

Define specific latent heat of vaporisation.

Answer 10

Specific latent heat of vaporisation is the quantity of heat required to convert unit mass of liquid to gas without any change of temperature.

Question 11

What are the differences between boiling and evaporation?

Answer 11

Both boiling and evaporation represent a change of phase from liquid to gas. Evaporation can take place at any temperature whereas boiling takes place at a fixed temperature.

Evaporation takes place at the surface of the liquid while boiling takes place throughout the liquid.

Question 12

Explain why a cooling effect accompanies evaporation.

Answer 12

Evaporation is the result of the exchange of energy between molecules. The molecules of a liquid are in constant random motion and they make frequent collisions with one another. During these collisions, some molecules gain energy while others lose energy. If a molecule near the surface of the liquid gains enough kinetic energy, it will be able to escape from the attractive forces of the molecules below it. Since the more energetic molecules escape, the average KE of the remaining molecules decreases, resulting in a decrease in the temperature of the liquid. Hence, evaporation results in cooling.

Question 13

State how the rates of evaporation can be increased.

Answer 13

1) Increasing the area of the liquid surface
2) Increasing the temperature of the liquid (i.e. increase the average microscopic KE of all the molecules)
3) Wind to remove the vapour molecules before they have a chance of returning to the liquid
4) Reducing the air pressure above the liquid (reduces the probability of a vapour molecule rebounding off an air molecule)

Question 14

Explain why the specific latent heat of vaporisation is higher than the specific latent heat of fusion for the same substance.

Answer 14

Firstly, to melt a solid, work must be done to separate some molecules against their mutual attractions so that the structure no longer has any rigidity. When a liquid vaporises, all the remaining bonds must be broken. Since melting means the breaking of relatively fewer bonds, we expect the specific latent heat of fusion to be less than that of vaporisation.

Secondly, the volume occupied in the gaseous state is much larger than that in the liquid state. The gas requires energy to do work against the external or atmospheric pressure (boiling must be done in an open container) during the expansion process.

Question 15

Outline the main principles in determining the specific heat capacity of soilids.

Answer 15

Solids (Pg 8 to 9 of notes) - heating a material through a change in temperature and comparing the electrical energy supplied

Question 16

Outline the main principles in determining the specific heat capacity of liquids and gases.

Answer 16

Liquids and Gases (Pg 9 to 10 of notes) - continuous flow method where a steady flow of fluid is passed along a pipe containig a heating coil. Measurements are only taken when the inlet and outlet temperatures have stabilised so as to ensure that the system has reached a steady state and the heat capacity of the apparatus can be ignored.

Electrical energy supplied = heat transferred to liquid + heat loss to the surroundings

By Newton’s Law of Cooling, the rate of heat loss H/t at the inlet and outlet is proportinoal to the excess temperature of the apparatus where Excess temperature of the body = Temperature of the body - Temperature of the surroundings

Hence, in order to eliminate H/t, the experiment has to be repeated with a different flow rate and adjusting the power of the heating coil such that the temperatures of the fluid at the inlet and outlet remain at the same values as that of the first experiment (assuming the temperatures of the surroundings remain unchanged).

Question 17

Outline the main principles in determining the specific latent heat of fusion.

Answer 17

A large funnel is filled with CRUSHED ice surrounding an electric heater. Before the heater is switched on, the apparatus is left until the water drips out of the funnel at a constant rate. A weighed beaker is then placed under the funnel so that the mass m of ice melted in a time duration t is determined. The heater is then switched on and the current is kept constant. When water drips out of the funnel at a constant rate, the mass M melted in time duration t is determined. If Lf is the specific latent heat of fusion for the solid, by principle of conservation of energy, electrical energy supplied + heat from surroundings = latent heat of fusion -> IVt + mLf = MLf

Question 18

Explain why a large amount of ice should be used at the start of the experiment determining the specific latent heat of fusion.

Answer 18

To ensure that the heater is always surrounded by ice.

Question 19

Outline the main principles in determining the specific latent heat of vaporisation.

Answer 19

An electric heater is immersed into a beaker of liquid that us placed on the pan or balance. the heater is switched on and the current us kept constant. When the liquid boils at a steady rate, the change in reading m on the balance in a time duration t is determined. If Lv is the specific latent heat of vaporisation of the liquid, by principle of conservation of energy, electrical energy supplied = latent heat of vaporisation + heat loss to surroundings -> IVt = mLv + H -> IV =mLv/t + H/t

Since the taste of heat loss H/t is proportional to the excess temperature of the body, the excess temperature would be the difference in temperature between the boiling point of the liquid and the temperature of the surroundings. In order to eliminate H/t, the experiment has to be repeated with a different boiling rate (m/t) by adjusting the heating power (since boiling point of liquid remains unchanged, the excess temperature of the liquid is the same).

Question 20

Define internal energy.

Answer 20

Internal energy of a system is the sum of the microscopic kinetic energy, due to the random motion of the molecules (leading to a range of K.E. where the average translational K.E. of one atom in the gaseous state equals to 3/2 kT) and the microscopic potential energy (always negative due to attractive forces), associated with the intermolecular forces of the system.

Question 21

State the relationship between temperature and internal energy. Explain why the internal energy of an ideal gas is directly proportional to its temperature l.

Answer 21

A rise in temperature implies an increase in the average microscopic kinetic energy of the molecules, which in turn implies an increase in the internal energy of the system, since internal energy is the sum of the microscopic kinetic energy and microscopic potential energy.

For ideal gases, there is negligible intermolecular forces between the molecules. Hence, its microscopic potential energy is 0 and its internal energy is only due to its translational microscopic kinetic energy and therefore directly proportional to its temperature.

Question 22

State the equations used to calculate the microscopic kinetic energy (KE) of ideal gases. Then, explain why any change in state will result in a change in the gas’ internal energy.

Answer 22

For a monoatomic ideal gas (i.e. gases made up of single atoms that only perform translational motion),

• microscopic KE of one molecule = 1/2 mc^2 = 3/2 kT
• microscopic KE of N molecules = 1/2 Nmc^2 = 3/2 NkT

From the ideal gas equation, pV = nRT = NkT.
U = 3/2 NkT = 3/2 nRT = 3/2 pV.

The state of s system is determined by its state variables pressure p, volume V and temperature T at that instant. Hence, any change will result in a change in internal energy U. (This is in contrast to work done)

Question 23

Explain why the factor of 3/2 only applies to monoatomic gases.

Answer 23

Monoatomic gases are made up of single atoms which only perform transitional motion only. Hence, 3/2kT is sometimes referred to as the mean translational kinetic energy. For non-monoatomic molecules, the factor 3/2 does not give the microscopic kinetic energy of each molecule. This is because the molecules now have rotational and vibrational motion, in addition to the translational kinetic energy.

Question 24

State the First Law of Thermodynamics.

Answer 24

The First Law of Thermodynamics states that the increase in internal energy of a system is equal to he sum of the heat supplied TO the system and the work done ON the system, and the internal energy of a system depends only on its state.

In equation form: INCREASE in internal energy of system = heat SUPPLIED TO the system + work done ON the system

When heat is transferred INTO the system/work is done ON the system (compression), internal energy increases. When heat is transferred OUT of the system/work is done BY the system (expansion), internal energy decreases.

By sign convention, positive work done BY gas = negative work done ON gas.

Question 25

Define heat.

Answer 25

Heat refers to the transfer of energy from one body to another, due to the temperature difference between the two bodies.

Question 26

Explain how work done on/by the gas can be calculated.

Answer 26

Pg 23 of notes

Integration is used as the pressure is constantly changing during the process.

Question 27

State the 5 special processes.

Answer 27

1) Isothermal Process (T is constant)
2) Isobaric Process (p is constant)
3) Isovolumetric/Isochoric/Isometric Process (V is constant)
4) Adiabatic Process (no heat transfer, Q = 0)
5) Cyclic Process (undergoes a series of changes before returning to original state)

Question 28

Differentiate between internal energy and work done.

Answer 28

Internal energy U is dependent only on the state of the system and the change in internal energy is hence independent of the path/process taken. However, work done on the other hand, is path dependent (depends on how it gets from its initial to final state).

For instance, comparing 3 processes, the change in internal energy can be the same. However, the work done/area under the p-V graph can be different.

Pg 25 of notes.

Question 29

Describe the Isothermal Processes, taking into account Q (heat transferred) and W (work done). Explain why the plot p-V shifts upwards at higher temperatures.

Answer 29

Pg 26 of notes

If the change in pressure and volume take place without a change in temperature, the process is known as an isothermal change. On the p-V graph, each line represents isotherms where all the points on one isotherm have the same temperature. Therefore, the change in internal energy U = 0. Applying the first law of thermodynamics where Change in U = Q + W, when U = 0, Q = -W.

During isothermal compression, volume decreases and work is done on the system. Therefore, W is positive and more than 0. For this process to occur, Q has to be transferred out of the system. Therefore, Q is negative.

During isothermal expansion, volume increases and work is done by the system. Therefore, W is negative and less than 0. For this process to occur, Q has to be transferred into the system. Therefore, Q is positive and more than 0.

Since pV = nRT, p = (nRT)1/V + 0. At higher temperature T, nRT increases and the plot of p-V shifts upwards.

Question 30

Describe the Isobaric Processes, taking into account Q (heat transferred) and W (work done).

Answer 30

Pg 27 of notes

If the gas undergoes heat exchange such that its volume change occurs at constant pressure, the process is known as an isobaric process.

During isobaric compression, volume decreases and work is done on the system. Therefore, W is positive and more than 0. Since pV = nRT, pV is proportional to T and Ta is less than Tb. Temperature decreases, causing internal energy to decrease and Change in U to be less than 0. Since U = Q + W and W is more than 0, Q must be less than 0. Therefore, heat is transferred out of the system

During isobaric expansion, volume increases and work is done by the system. Therefore, W is negative and less than 0. Temperature increases, causing internal energy to increase and Change in U to be more than 0. Since U = Q + W and W is less than 0, Q must be more than 0. Therefore, heat is transferred into the system

Question 31

Describe the Isovolumetric/Isochoric/Isometric Process, taking into account Q (heat transferred) and W (work done).

Answer 31

If the change in the pressure of the gas occurs without a change in volume, the process is known as an isovolumetric process. Since V is constant, there is no compression or expansion. Thus, W = 0. Applying the first law of thermodynamics where Change in U = Q + W, since W = 0, then Change in U = Q. If p increases, the temperature will increase and U increases. Hence, heat is transferred to the system as both the change in U and Q will be positive. However, if p decreases, the temperature will decrease and U decreases. Hence, heat is transferred out of the system as both the change in U and Q will be negative.

Question 32

Describe the Adiabatic Processes, taking into account Q (heat transferred) and W (work done). Explain how such as process can be attained.

Answer 32

Pg 29 of notes

When the change in pressure and volume occurs with no heat transferred into and out of the system, it is known as an adiabatic process. Such a process can be attained when the system is insulated or when the change in pressure and volume occurs faster than the exchange of heat with the surroundings (i.e. the process occurs rapidly; e.g. the bursting of a balloon). A change in volume results in a change in temperature along with a change in pressure.

No heat transfer implies that Q = 0. Since the change in internal energy U cannot be 0, applying the first law of thermodynamics where Change in U = Q + W, since Q = 0, then Change in U = W. During adiabatic expansion when volume increases, work is done by the system and work done on the system is negative. Pressure decreases and temperature decreases. Therefore, the change in U is negative and W is also negative.

During adiabatic compression, when volume decreases, work is done on the system and work done on the system is positive. Pressure increases and temperature increases. Therefore, the change in U is positive and W is also positive.

Question 33

Describe the Cyclic Processes, taking into account Q (heat transferred) and W (work done).

Answer 33

A cyclic process is one in which the system goes through a series of processes and ends at its initial state. Therefore, the change in internal energy U is 0. Applying the first law of thermodynamics where Change in U = Q + W, since U = 0, then Q = -W. By comparing the area under the p-V graph, the next work done on the system can be found. This work done is also equal to the area enclosed by the cycle. For this process, there is net heat transfer out of the system or heat net heat lost from the system.