# Questions for Thermal Physics (JC) Set 2

Show/Hide Sub-topics (Thermal Physics | A Level)

a) (i) The First Law of Thermodynamics is in agreement with the Principle of Conservation of Energy. Give a word equation stating the First Law of Thermodynamics.

(ii) A student suggests that when heat is supplied to a system, the temperature will always increase. Give an example to show when the statement is not true.

b) When a system is taken from state a to state b in the diagram below along the path acb, 90.0 J of heat flows into the system and 60.0 J of work is done by the system.

i) Show that the increase in internal energy for path acb is 30 J.

ii) Explain why the increase in internal energy for path acb is the same as adb.

iii) Hence calculate the amount of heat that flows into the system for path adb if 15.0 J of work is done by the system.

iv) If the increase in internal energy from a to d is 8.0 J, find the amount of heat absorbed in the process db.

v) When the system is returned from b to a along the curved path, the value of the work done on the system is 35.0 J.

State whether the system gained or lost heat and calculate this amount of heat exchanged with the system?

c) i) State one assumption that can be made when a gas is assumed to be ideal.

ii) A rubber sphere has a fixed internal volume of 0.0120 m3 and is kept at a temperature of 25°C. What is the amount of air in moles, that need to be supplied into the sphere to increase its pressure from 2.62 x 105 Pa to 3.23 x 105 Pa, without a change in temperature? (You may assume that air is an ideal gas.)

iii) Show that the internal energy of one molecule of air at a temperature of 25°C is 6.17 x 10-21 J. Assume that the air behaves as a mono-atomic ideal gas.

iv) Hence, calculate the increase in internal energy of the air in the rubber sphere as a result of the increase in pressure at constant temperature.

a) i) The First Law of Thermodynamics states that internal energy is a function of state and the increase in internal energy of a system is equal to the sum of the heat supplied to the system and the work done on the system.ii) Examples: Object expands at constant temperature when heat is supplied OR Answer stating changes of phase (melting or boiling.b)
i) ΔU = Q + WD
= 90 + (-60)
= 30.0 Jii) Internal energy is a function of state (path independent). OR Initial state and final state are the same for both paths.iii) ΔU for path acb is same for path adb (ΔU is path independent)
hence, 30 = Q + (-15)
Therefore Q = 45 Jiv) From a to d to b
30 = 8 + ΔUdb
ΔUdb

From d to b,
1st Law: ΔU = Q + WD
22 = Q + 0
Q = 22 J

v) From a to b, ΔU = 30 (increase in internal energy)
From b back to a, ΔU = -30 (decrease in internal energy)

1st Law: ΔU = Q + WD
-30 = Q + 35
Q = -65 J

65 J of heat is released.

c) i) Any one of these

• intermolecular forces between molecules is negligible
• collisions between molecules (and with walls of container) are elastic
• volume of molecules negligible compared to volume occupied by gas

ii) PV = nRT
Hence, Δn = ΔPV / (RT)
= (3.23 – 2.62) x 105 x 0.0120 / (8.31)(25 + 273.15)
= 0.30 mol

iii) U = 3/2 NKT
= 3/2 (1)(1.38 x 10-23)(25 + 273.15)
= 6.17 x 10-21

iv) Total U = internal energy of 1 molecule x number of moles x Avogadro’s constant
= (6.17 x 10-21)(0.30)(6.02 x 1023)
= 1.1 x 103 J

The specific latent heat of vaporisation of water at 20°C is appreciably greater than the value at 100°C. This is because

1. the specific latent heat at 20°C includes the energy necessary to raise the temperature of one kilogram of water from 20°C to 100°C.
2. more work must be done in expanding the water vapour against atmospheric pressure at 20°C than at 100°C.
3. the molecules in the liquid are more tightly bound to one another at 20°C than at 100°C.
4. the root mean square speed of the vapour molecules is slower at 20°C than at 100°C.

Specific latent heat of vaporisation is the amount of energy to change a substance from liquid phase to gaseous phase without a change in temperature. Therefore, it is more difficult to have a phase change at 20°C compared to that at 100°C as the molecules in the liquid at 20°C are more tightly bound to one another.

An ideal gas, contained in a cylinder by a frictionless piston, is allowed to expand from volume V1, at pressure p1, to volume V2, at pressure p2. Its temperature is kept constant throughout. The work done by the gas is

1. positive because the volume has increased
2. zero because intermolecular forces are negligible in an ideal gas.
3. zero because it has been kept at constant temperature and so its internal energy is unchanged.
4. negative because the pressure has decreased and so the force on the piston has been diminishing.

During expansion at constant temperature, the work done on the gas is negative as the gas expands. Therefore, the work done by the gas is positive.

a) State two quantities which increase when the temperature of a given mass of gas is increased at constant volume.

b) A car tyre of volume 1.0 x 10-2 m3 contains air at a pressure of 300 kPa and a temperature of 17°C. The mass of one mole of air is 2.9 x 10-2 kg.

Assuming that the air behaves as an ideal gas, calculate

i) n, the amount, in mol, of air
ii) the mass of air
iii) the density of the air

c) Air containing oxygen and nitrogen molecules. State, with a reason, whether each of the following quantities are the same for oxygen and nitrogen molecules in air at a given temperature.

i) Average kinetic energy
ii) Root mean square speed

a) Any two of the following:
Pressure, Average kinetic energy, Root mean square speed, internal energyb) i) pV = nRT
n = {[(1.0 x 10-2)(300 x 103)]/(8.31 x 290)}
n = 1.24 molii) Mass of air = 1.24 x 2.9 x 10-2
Mass of air = 3.6 x 10-2 kgiii) Density = (3.6 x 10-2) / (1.0 x 10-2)
Density = 3.6 kg m-3c)
i) Same because they are at the same temperature.
ii) Different because the mass of the molecules are different.

Two containers of volume 4.0 m3 and 6.0 m3 contain an ideal gas at pressure 100 Pa and 50 Pa respectively. Their temperatures are equal. They are joinged by a tube of negligible volume. The gas flows from one container to the other with no change in temperature. The final pressure will be

1. 70 Pa
2. 75 Pa
3. 80 Pa
4. 150 Pa

Initially, na = PaVa / RTa, nb = PbVb / RTb
Final: naf = PfVf / RTf

Since na + nb = naf + nbf
PaVa + PbVb = PfVf
100(4) + 50(6) = Paf(6 +4)
Pf = 70 Pa.