Physics JC1 2009

Assignment 6: Thermal Physics

1(a)(i)Explain what is meant by the terms

Internal energy of a gas,

Ideal Gas[2]

(ii)State, in words, the relation between the increase in the internal energy of a gas, the work done on the gas, and the heat supplied to the gas. [1]

Solution

(a)(i)Internal energy of a system is the sum of a random distribution of microscopic kinetic energy and potential energy of all the molecules in the system. It is dependent on its state. [1]

An ideal gas is a gas that obey the equation of state, pV = nRT, for all values of pressure, volume and temperature. [1]

(ii)The increase in the internal energy of a system is the sum of the work done on the system (gas) and the heat supplied to the system (gas). [B1]

(b)The gas in the cylinder of a diesel engine can be considered to undergo a cycle of changes of pressure, volume and temperature. One such cycle, for an ideal gas, is shown on the graph.

The temperature of the gas at A and B are 300 K and 660 K respectively. Use the ideal gas equation and data from the graph to prove that the temperatures at C and D are 2830 K and 2300 K respectively. [2]

Solution

bpV = nRT

For constant amount of gas, n is constant

[1]

Tc = 2830 K

[1]

TD= 2300 K

(c)(i) During each of the four sections of the cycle, changes are being made to the internal energy of the gas. Some of the factors affecting these changes are given in the table below.

1.Explain why the work done on the gas is sometimes negative and find the work done on the gas in section D to A. [2]

2. Deduce the values of the “increase in internal energy of the gas” for each section and list them. [2]

Section of cycle / Heat supplied to gas/ J / Work done on gas/ J / Increase in internal energy of gas/ J
A to B / 0 / 300
B to C / 2580 / – 740
C to D / 0 / – 440
D to A / – 1700

(ii)Explain why the total change in the internal energy of the gas during a complete cycle must be zero. [2]

Solution

c(i)1.Work done on gas is negative because the gas expands. [1]

Work done on gas from D to A is zero [1], as there is no change in the volume.

2.

Section of cycle / Heat supplied to gas/ J
Q / Work done on gas/ J
W / Increase in internal energy of gas/ J
(ΔU = Q + W)
A to B / 0 / 300 / (0) + (300) = 300
B to C / 2580 / – 740 / (+2580) + (–740) = 1840
C to D / 0 / – 440 / (0) + (–440) = –440
D to A / – 1700 / 0 / (–1700) + 0 = –1700

(2 – remove one mark for each wrong entry, max is 2)

(ii)The internal energy of the system depends on the state of the gas[1] (the state of gas refers to its pressure, volume and temperature if the mass of gas used is a constant). Since the gas after one cycle is back at the same pointon the p-V graph (i.e. the starting and ending point of the process is the same)[1], the internal energy remains unchanged.

2(a)For each of the following statements below, discuss whether it is correct, stating clearly your reasons.

(i)The latent heat of fusion of a substance is always less than the specific latent heat of vaporisation of the substance. [2]

(ii)For the same mass, volume and temperature, the pressure exerted by a real gas is the same as the pressure exerted by an ideal gas. [2]

Solution

(a)(i)During melting of solid, the forces of attraction between the molecules are merely weaken whereas the forces of attraction between the molecules are totally broken during vaporization of a liquid. Thus the molecules are less further apart in the case of vaporization. Hence, the increase in potential energy during melting is less than the increase during vaporization.

In addition, work is also done against external pressure as the volume of the substance increases.

 True

(ii)Pressure of a real gas is less than that of an ideal gas for the same mass, volume and temperature. This is because attractive forces exist between the atoms or molecules in a real gas and not in an ideal gas.

 False

(b)A styrofoam container of negligible heat capacity contains 900 g of water at 85 C. 100 g of ice at 0 C was then added to water.

Specific latent heat of fusion of water /Jkg–1 / 3.34 × 105
Specific heat capacity of water /Jkg–1K–1 / 4200
Specific latent heat of vaporisation of water /Jkg–1 / 2.26 × 106

(i)Calculate the equilibrium temperature of the mixture. State one assumption made in your calculation. [3]

(ii)Explain why exposure to steam at 100 C produces a more severe burn than exposure to the same amount of hot water at 100 C. [2]

Solution

(b)(i)Assuming no heat lost to surroundings,[1]

Heat supplied by water = heat gained by ice

[1]

Θ = 68.5 °C[1]

(ii)Consider the same mass of steam m and the same mass of water m at 100 C. When in contact, both steam and skin reaches a thermal equilibrium. The same happens for boiling water with skin. However, the difference is that for steam, the process of condensation has to occur[1] prior to the drop in temperature.

When steam condenses into water at 100 C, there is a release of a large amount of thermal energy. Since specific latent heat of vaporisation is far greater than specific heat capacity [1], the thermal energy transferred to the skin due to steam condensing is greater than that due to boiling water.

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