PHAS0006 Thermal Physics and the Properties of Matter Solutions Exam 2021
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PHAS0006
Thermal Physics and the Properties of Matter
Exam 2021
1. (a) Two thermally isolated objects of mass m1 and m2 are initially at temperatures T1 and T2 . Each mass has specific heat capacity c1 and c2 .
(i) Find an expression for the final temperature Tf when the two objects are brought into thermal contact and have come to thermal equilibrium. Round these to the nearest K.
Soln: Q1 = m1 c1 ( Tf - T1 ) and Q2 = m2 c2 ( Tf - T2 ) and Q1 = -Q2 .
Solving for Tf gives Tf = [3U].
(ii) Calculate the final temperature at equilibrium when a 100 kg block of ice, at a temperature of 200 K, is brought into contact with a 50 kg block of concrete at a temperature of 180 K. The specific heat capacity of ice is 1.34 kJ kg_1 K_1 , and for concrete is 1.84 kJ kg_1 K_1 . Assume that the two blocks are thermally isolated and that the specific heat capacities do not change with temperature.
Soln:
Tf = =191.9 K[2U].
(iii) Determine the change in entropy when the ice and concrete have come to equilibrium using the values given in (ii).
Soln:
Entropy change when heat is transferred from hot to cold ∆S1 = m1 c1 /TdT = m1 c1 ln Tf /T1
and for the cold to hot ∆S2 = m2 c2 /TdT = m2 c2 ln Tf /T2 . The total change is
∆S = 100*1340 ln 191.9/200+50*1840 ln 191.9/180=349.65 J/K [3U].
(b) A thermally isolated container consists of a cylinder with a movable piston as shown in the diagram below. The container has a total mass m1 , specific heat capacity c1 and is initially at a temperature T1 . n moles of an ideal, monatomic gas, initially at a temperature of T2 , are introduced into the container via an adiabatic free expansion.
(i) Find an expression for the final equilibrium temperature of the container and the gas inside it at fixed volume. Assume that they are thermally isolated from the environment. Soln:
Q1 = m1 c1 ( Tf - T1 ) as before but now Q2 = 3/2nR( Tf - T2 ) for an ideal gas. For Q1 = -Q2 , the final temperature is now
Tf = [3U]
(ii) Using your result from part (i), or otherwise, determine the final equilibrium temperature when the container has a mass 10 kg and a specific heat capacity of 2 kJ kg_1 K_1 . The container is initially at a temperature of 300 K and there are
10 moles of gas initially at a temperature of 200 K. Soln:Tf = =299.4 K [2U]
(iii) Calculate the gas pressure when the container volume is 0.01 m3 . Soln:p = nRT/V = (10 * 8.3145 * 299.4/0.01) = 2.49 MPa [1U]
(iv) The gas is now allowed expand and the volume increases [2]
by 0.01 m3 . When this isolated system comes to equilibrium, does the temperature change? What type of thermodynamic process is this?
Soln:No the temperature does not change as it is a closed system. It’s an isothermal process[2U].
(v) Determine the work done by the gas and the change in its entropy for the thermodynamic process considered in (iv). Soln:
For an isothermal process W = -nRT ln Vf /Vi =
10 * 8.3145 * 299.4 ln (0.02/0.01) = 17.3kJ [2U]
The change in entropy for an ideal gas is
∆S = 3/2nR ln U/U0 + nR ln V/V0 . As it is an isothermal process and U = U0 , ∆S = 10 x 8.3145 ln (0.02/0.01) = 57.6 J/K [2U].
2. Considering both Carnot and Otto engine cycles: (a) Describe with the aid of a pV diagram, the thermodynamic processes that occur in each cycle when the working fluid is an ideal monatomic gas. Indicate in which part of each cycle heat is transferred to and rejected by the fluid.
Soln:
0.5 mark for each of the four processes in each diagram [4S] 0.5 mark for each of two heat transfer processes in each diagram [1S]
(c) The maximum temperature reached in an Otto cycle is 1000 K, and the minimum temperature is 350 K. The heat transfer into the engine is 4 J per cycle and 4 J is rejected. Calculate the two intermediate temperatures reached within this ideal cycle for 0.001 moles of gas.
Qin = CV ( Tmax - T1 ) and Qout = CV ( T2 - Tmin ) [1S]
T1 = Tmax - Qin /CV , T2 = Qout /CV + Tmin and CV = 3/2nR [1U] T1 = 1000 - 4/(1.5x.001x8.3145) = 679.28 and
T2 = 350 - 2/(1.5x.001x8.3145) = 670.74 [1U] (c) Describe how heat is transferred to and rejected by the working fluid in an internal combustion engine. What practically limits the efficiency of this type of engine?
Soln: Heat is transferred by combustion of fuel-air mixture and heat is rejected into the engine block [1S] The maximum compression ratio defined by auto-ignition of fuel-air mix [1S]
The Lennard-Jones interaction potential can be written as:
V(r) = 4e ┌ ╱ 、 12 - ╱ 、 6 ┐ .
(d) Discuss briefly what this function describes, and explain the physical origin of the two terms in the square brackets. Soln: This function describes the potential energy between two atoms a distance r apart [1S].
Soln: The term involving r_12 arises from short-range repulsion due to overlap of core electrons/Pauli exclusion principle [1S]. The term involving r_6 arises from longer-range attraction due to fluctuating/induced dipole interactions[1S].
(e) With the aid of a diagram(s), explain the microscopic origin for the thermal expansion of simple solids.
Soln: Diagram something like above – for [1S]. Doesn’t have to be perfect – just showing an asymmetric well or the interatomic potential, V(r). Explanation: the interatomic potential energy/well is not harmonic: i.e. it is not symmetric about equilibrium separation/about r0 /rises steeper at low-r than high-r [1S]. As the kinetic energy of particles increases,
the mean/average spacing between particles/atoms therefore increases [1S].
(f)(i) A cylindrical aluminium bar has length 2 m and radius 0.2 m, and is at equilibrium at 300 K. What is the length of the bar at 350 K? The linear expansion coefficient of aluminium is 2.3 x 10_5 K_1 .
Soln: ∆L = αL∆T[1S]= 0.0023, therefore new length is 2.0023 m. [1U]
(ii) If the aluminium bar is held so that it cannot expand, what pressure is exerted on the ends of the bar when the temperature is increased from 300 to 350 K? The Young’s modulus for aluminium is 6.9 x 1010 Pa.
Soln: = E [1S]
= p = 6.9 x 1010 x 0.0023 = 158700000 Pa [1U]
3. (a) Explain two cases where point defects can be beneficial to the functional properties of a material, and two cases where point defects can be detrimental to the functional properties of a material.
Soln: GOOD - NV centres, Mg in Al drinks cans, C in Fe for steel, gem stones, malleability of metals etc.BAD, work hardening and brittlisation of metals, loss of conductivity etc., lack of clarity in GEMs etc. For each of the four cases [0.5] for identifying the examples and [0.5] for brief explanation
(b) In a diffraction experiment on a cubic crystal of unit cell length 4.3 A˚ , at what 2θ values would you find the 111 and the 220 reflections, using X-rays of wavelength 0.154 nm?
Soln: sinθ = ; θ = sin_1 ╱ 、; d = [1S] a = 0.43 nm; λ = 0.154 nm
1 1 1; d = ; θ = sin_1 ╱ 、= 18.07 。; 2θ = 36.14 。 [1U] 2 2 0; d = ; θ = sin_1 ╱ 、= 30.43 。; 2θ = 60.86 。 [1U]
(c) A cubic ionic crystal has a chemical composition AB. If ion B has a radius of 0.15 nm, calculate the smallest radius of ion A for which AB has a crystal structure with eight nearest neighbours and the A ions touch the B ions.
Soln: 8 nearest neigbours means it’s a cubic crystal with ion B at 0,0,0 and ion A at 1/2,1/2,1/2. [1.5U]
Smallest radius of middle ion is when the corner ions touch.
unit cell length a = 2 * 0.15 [0.5U]
Central ion radius is therefore calculated by 1/2 * cube diagonal-a [0.5U]
r = ((A3 x a) - (a))/2 = 0.11 nm [1.5U]
(d) By sketching isotherms of the van der Waals equation of state at temperatures above and below the critical point, explain Maxwell’s equal area rule and its physical implications with reference to experimental observations.
Soln:
Soln: Graph axis [0.5S], Shape [0.5S], straight line drawn bisecting equal areas [1S]
Maxwell’s equal area construction is performed by drawing a striaght line (i.e. a line of constant pressue) across the unphysical region in the pV isotherm, so that the areas above and below this line are equal [1S] as shown in the diagram. With the line drawn as such, the isotherm now replicates experimental data. The change in volume at constant pressure is explained by the fact that the gas and liquid phase can co-exist [1S]
(e) The vapour pressure of carbon dioxide increases from pvap to 1.5pvap when the temperature increases by 20 degrees Kelvin. The enthalpy of vaporisation of carbon dioxide is 15.3 kJ mol_1 . Calculate the value of the temperature before it was increased.
Soln:
We require the Clausius-Clapeyron equation.
ln = ╱ - 、[1.5U]
ln(1.5) = ╱ - 、[0.5U]
ln(1.5) = [0.5U]
T2 + 20T - = 0 [1U]
T = _20oA202 +4×1×(90801..)
2
T = 291.5 K [1U]
2022-04-22