Structure of Matter 23.03.11 First Year Physics Classwork ...

Structure of Matter 23.03.11First Year Physics

Classwork IInternal Combustion Engines

Information needed for this Classwork

• PV = nRT ideal gas equation of state (R ≃ 8.31 JK−1mol−1)

• For gas, dW = −PdV

• For adiabatic changes,PV γ = constant, where for a fuel mixtureγ ≃ 1.4

• At constant volume, dQ = nCV m dT

1. A petrol engine (Otto cycle) can be represented by a 6 processcycle (see figure below);

(a) In the intake stroke, the piston moves to allow the air/fuel mixture to enter the chamber at closeto atmospheric pressure (isobarically).

(b) The intake valve is then shut, and the mixture is compressed adiabatically.

(c) Once compressed, the spark plugs fire to ignite the mixture. This results in an almost instanta-neous rise in pressure at a constant volume (isochorically)

(d) The heated fuel is allowed to expand adiabatically, until the chamber is once more at its maxi-mum volume.

(e) Whilst the chamber is at its maximum volume, the exhaust valve is opened and there is a de-compression as hot gas leaves the chamber.

(f) Finally, in the exhaust stroke, the remaining spent fuel is expelled with the exhaust valve open toatmospheric pressure. The exhaust valve is then shut and the intake valve opened in preparationfor a new cycle.

Figure 1: 6 stages of petrol engine

Draw this cycle on aPV diagram. What represents the work done by the engine per cycle on thediagram?

2. If the maximum volume of the cylinderV0 = 0.5 litres (= 0.5×10−3 m3), if it is filled at atmosphericpressure (∼ 105 Pa) and27 ◦C. Calculate the number of moles of fuel mixture, assuming it is allgaseous.

3. Use the adiabatic relationPV γ = c and the ideal gas equation of state to find a relation between thevolume and temperature in an adiabatic process. Hence for a compression ratio of 10:1, calculate thepressure and temperature(P1, T1) after the compression phase.

4. Show that the work doneon a gas during an adiabatic compression from a state with temperatureT0

to one with temperatureT1 is given by,

∆W =nR

γ − 1(T1 − T0)

5. Calculate (numerically) the work done by the compression detailed in part 3. Comment on the signof the work.

6. If the mixture’s average molecular mass isA∗≃ 35 and has a molar heat capacityCV m = 5

2R, and

the energy of combustion isHc ≃ 500 kJ kg−1, calculate the temperature change on combustion.(Assume the number of moles of gas remains constant).

7. Calculate the temperature change and hence work done on expansion. What is the total work done bythe gas during the cycle.

8. If an engine with four such cylinders is running at around 3000 revolutions per minute, calculate thepower output. (You may want to quote this in horsepower, 1 hp≈ 746 watts).

9. A diesel engine works in almost the same way as a petrol engine, except that the fuel is compressedto much higher pressure and temperature so that it self-ignites instead of requiring a spark for com-bustion. This burn phase can be represented as an isobaric expansion at the same final pressure asattained in the petrol engine. Include thePV cycle for a diesel engine of the same volume as yourpetrol engine on the same diagram. Which would be more efficient? Can you say why petrol enginesare more common in cars?

Numerical Answers:2) 0.02 moles 3)2.5 × 106 Pa, 754 K 5) 189 J 6) 840 K 7) -960 K, -399 J, 210 J 8) 21 kW, 28 hp


Classwork IIThe Isothermal Atmosphere


• Probability that a given molecule (massm) is between heights z and z+dz in an isothermalatmosphere (temperature T ): (mg/kBT )e−mgz/kBTdz.

• Boltzmann’s constant: kB = 1.38× 10−23 J K−1.

• Absolute zero = −273◦C.

• Acceleration due to gravity: g = 9.81 m s−2.

• Radius of earth rE = 6, 378 km

• Height of Mount Everest = 8.8 km

• For air, adiabatic constant γ = 1.4 and average molecular mass m = 29× 1.66× 10−27 kg

1. The probability that a given molecule in an isothermal atmosphere is between heights z

and z+ dz can be written p(z)dz, where p(z) =1

λe−z/λ. Write down an expression for the

‘scale height’ λ in terms of T , the temperature, and m, the mass of a molecule.

2. The probability that a molecule is located between z = z1 and z = z2 is∫ z2

z1p(z) dz.

(a) Calculate the probability that a given molecule is between z = 0 and z = λ.

(b) Calculate the probability that a given molecule is between z = λ and z = 2λ.

3. Show that∫ ∞0

p(z) dz = 1, and interpret this result.

4. Calculate λ for T = 20◦ C and m = 4.82 × 10−26 kg (the average mass per molecule inair).

5. There is a 99.9% probability that any given molecule is located below a certain height z∗.Calculate the value of z∗ for the parameters given in Q 5.

6. Write down a corresponding expression for the pressure as a function of height using thismodel, assuming atmospheric pressure at ground level is 1 bar. If humans can survive atpressures down to 0.5 bar, calculate the maximum height for human habitation.

7. In deriving the isothermal atmosphere model we assumed that g, the acceleration due togravity, was independent of height. Was this a reasonable assumption?

8∗. Of course, the earth’s atmosphere is not isothermal. Better (though still not perfect) wouldbe to assume an adiabatic relation; i.e. that as air rises, it does not exchange heat withneighboring air and so cools. Starting with the ideal gas equation of state, the first law foran adiabatic process and the theorem of equipartition of energy, show that the following

relationship holds; dP =γ

γ − 1

(P

T

)dT

9∗. Use this and the expression for the pressure differential for a slab in equilibrium to constructa differential equation for the rate of change of temperature with height. Hence show thatthe adiabatic lapse rate (the rate at which temperature falls as a function of height) is≈ 9.8◦Kkm−1.

10∗. What would be the temperature at the top of Mount Everest, assuming a temperature of20◦C at ground level. By solving the differential equation in part 9∗, find the pressure atthis height too, and compare it to the value obtained from the isothermal model.

Numerical Answers2a) 0.63, b) 0.23; 4) 8.55 km; 5) 59.1 km 6) 5.9 km 10) Padiab = 0.29 bar; Pisoth = 0.35 bar


Classwork IIIDerivation of the Ideal Gas Equation of State

Figure 1: a collision

In Lecture 1 we derived the ideal gas equation of state by considering gasmolecules undergoing elastic collisions with a wall of the container and im-parting momentum to it. The pressure is equal to the rate of momentumtransfer per unit area. However, one aspect of that derivation was highlyunsatisfactory. In this Classwork we do it properly.Consider a box of volume V containing N molecules. Define the x axispointing into the wall, as shown. The dodgy part of the previous deriva-tion was the assumption that all the molecules had the same value of |vx|,and then at the end replacing |vx|2 with the average value without any realjustification. But we know that the molecules have values of vx between−∞ and +∞, according to the velocity component distribution function: f(vx) = Ae−αv

2x where

A =

(m

2πkBT

)1/2

and α =m

2kBT.

1. Write down an expression for the total number of molecules in the volume V with vx betweenvx and vx + dvx, and, hence, show that the number density of such molecules (the number perunit volume) is nf(vx)dvx where n = N/V .

2. Considering particles with 0 < vx < +∞ (molecules with vx < 0 are moving away from thewall and won’t hit it), show that the number of impacts with area L2 of the wall1 in time ∆t bymolecules with vx between vx and vx + dvx is dNcoll = n vxf(vx) dvx ∆t L2.

3. In Sec. 1.1 we foundNcoll = 12L2|vx|n∆t. Why doesn’t the factor 1

2appear in the new equation

for dNcoll, in Q. 2?

4. By integrating the expression for dNcoll in part 2 (using the considerations in part 3 to chooseappropriate limits), show the rate that particles hit the wall per unit surface area is given by

14n〈v〉, where 〈v〉 ≡

(8kBT

πm

)1/2

is the mean speed.

5. Each impact imparts momentum 2mvx to the wall. Show that the contribution to the pressureon the wall from the molecules between vx and vx + dvx is dP = 2nmv2xf(vx)dvx.

6. Using the standard integral∫ ∞0

x2e−αx2

dx =1

4

( πα3

)1/2show that P = nkBT (the ideal gas

equation of state).

7. It is tempting to write the pressure (≡ total momentum change on the wall per unit area) to bethe average rate at which particles strike the wall (part 4) multiplied by a change of momentumper collision of 2m〈v〉. Calculate this quantity (in terms of T ). Can you give reasons why thisis not the same as the expression given in part 6.

1In Sec. 1.1 we considered an area A of the wall, but here we use the symbol A for the normalising constant in f(vx).


Classwork IVSurface Tension


• 1 atmosphere = 1.01× 105 N m−2.

• Acceleration due to gravity: g = 9.81 m s−2

• Density of water: ρ = 103 kg m−3

• Surface tension of water: γ = 0.0728 N m−1

Liquids try to minimize their surface area; a liquid surface islike a stretched elastic membrane. The force responsible forthis is surface tension.

Consider a soap film on a wire frame, as shown. Thesurface to the left of the imaginary line AB pulls theline to the left, but the surface to the right exerts anequal force to the right. The force acts perpendicularto the line and is proportional to the length, l, of theline; i.e. F = γl, where γ is the surface tension (unitNm−1).

1. Assume section CD of the wire frame is movable. To keep it in place a force F0 must beapplied that is equal to the surface force pulling to the left. Increasing the applied force aninfinitesimal amount dF will result in section CD moving an infinitesimal distance dx to theright. Show that the work done is 2γl dx. Hence, show that the increase in energy of thesoap film can be written γ dA, where dA is the change in the surface area. (Hint: remembera soap film has two surfaces.)

2. It requires energy γ dA to increase the area of a fluid surface. You might think that at themicroscopic (i.e. atomic) level this energy is used to pull the atoms further apart and increasetheir separation. But, in fact, the average interatomic separation doesn’t change; liquids areincompressible, so the volume of the soap film is constant. The thickness of the film simplydecreases when the area increases.

So, at the microscopic level, where does the energy go?

3. The surface of a spherical bubble (of radius r) in a liquid (e.g. fizzy drink) experiences a netinward force Fc towards the centre. This force can be found by calculating the work neededto increase the size of the bubble. If an infinitesimally larger outward force is applied to thesurface the radius will increase by dr and the work done will be Fc dr. Find an expressionfor dA, the increase in surface area corresponding to an increase in radius of dr, and henceshow that Fc = 8πγr. (Hint: remember that this type of bubble, more properly called acavity, has only one surface.)

4. In addition to this surface tension force, the surface is also acted on by Pout, the pressureof the surrounding liquid, and Pin, the pressure of the gas inside the bubble. Show that

Pin = Pout +2γ

r. (This equation is called the Laplace equation.)

5. It is proposed to aerate a large open tank of water by pumping air into the water at a depthof 1 m below the surface. What air pressure is required to produce bubbles of radius 0.1mm. Assume that the pressure at the surface of the water (i.e., at the top of the tank) is 1atmosphere.

6. If a liquid is sufficiently pure (without bubbles or contaminants), and stored in a very smoothcontainer, it is possible to heat it to well above its boiling point before it begins to boil. Bylooking at the form of the Laplace equation, suggest why it is more difficult to bring a pureliquid to boil.

Numerical Answer5) 1.12× 105 N m−2


Classwork I - AnswersInternal Combustion Engines

1. PV diagram for Otto cycle:

2. FromPV = nRT , n =PV

RT=

105∗ 5 × 10−4

8.31 ∗ 300≃ 0.02 moles

3. P1Vγ1

= P0Vγ0

So P1 = P0

(

V0

V1

)γ

= 1 atm. ∗

(

V0

(V0/10)

)1.4

= 101.4atm = 25.1 atm. ≃ 2.5 × 106 Pa

Also PV γ =

(

nRT

V

)

V γ so TV γ−1 = constant

→ T1Vγ−1

1= T0V

γ−1

0

T1 = T0

(

V0

V1

)γ−1

= 300

(

V0

(V0/10)

)

1.4−1

= 300 ∗ 100.4≃ 754 K

4. ∆W =

∫ P1,V1

P0,V0

−P dV but P = CV −γ

So, ∆W =

∫ V1

V0

−CV −γ dV =

[

−1

1 − γCV −γ+1

]V1

V0

→ ∆W =

[

1

γ − 1CV −γ V

]V1

V0

=

[

1

γ − 1PV

]P1,V1

P0,V0

but for ideal gas,PV = nRT , so

→ ∆W =

[

1

γ − 1nRT

]T1

T0

→ ∆W =nR

γ − 1(T1 − T0)

(Alternatively starting from dU = dQ + dW = dW for adiabatic process since dQ = 0,

butU =nd

2nRT → dU = dW =

nd

2nRdT (= CV dT )

But γ =nd + 2

nd

→ nd(γ − 1) = 2 →

nd

2=

1

γ − 1

So∆W =nR

γ − 1∆T =

nR

γ − 1(T1 − T0))

5. Work done (from above)=nR

γ − 1(T1 − T0) =

0.02 ∗ 8.31

0.4(754 − 300) = 189 J.

The sign should be positive to indicate work is done on compressing the gas.

6. Mass of gasM = mn10−3 kg = 35 × 10−3∗ 0.02 = 7 × 10−4 kg

So heat of combustion∆Q = Hc ∗ M = 500 × 103∗ 7 × 10−4 = 352 J

But ∆Q = C∆T = CV m ∗ n ∗ ∆T (since heating is at const.V )

∴ ∆T =∆Q5

2Rn

=350

2.5 ∗ 0.02 ∗ 8.31= 840 K

andT2 = T1 + ∆T = 754 + 840 = 1594 K

7. usingT3Vγ−1

3= T2V

γ−1

2, whereV2 = V1 = V0/10 andV3 = V0

So T3 = T2

V0/10

V0

0.4

= T210−0.4 = 634 K

So∆T = 634 − 1594 = −960 K

and ∆W3 =nR

γ − 1(T1 − T0) =

0.02 ∗ 8.31

0.4(−960) ≃ −399 J

(NB negative this time, since work done by gas as it expands, and so negative work done on it!)

Since no work is done during isochoric processes, total work done= ∆W1 + ∆W3 = 210 J

(NB now +ve since work doneby engine = work doneby gas).

8. This is the work done in one cylinder in one cycle, but for four cylinders and each cycle takes 2revolutions (think about how many compression and expansions are required per cycle), so rate ofcycles = 0.5 * 3000 rpm = 25 Hz

Power output= ∆W ∗ 25 ∗ 4 = 210 ∗ 100 watts = 21.0 kW = 28hp

9. Diesel engine gives greater shaded area and∴ greater work per cycle, and so is more efficient (greaterwork for same amount of heat of combustion).

However, since cycle takes longer, diesel engines cannot be revved so fast, and therefore typicallyproduce less power for the same size. Also to withstand higher compressions, they have to be heavier,and are often restricted to heavier vehicles (like trucks...).


Classwork II - AnswersThe Isothermal Atmosphere

1. Comparing (mg/kBT )e−mgz/kBTdz and p(z) =1

λe−z/λ gives λ = kBT

mg

2a.∫ λ

0p(z) dz=

[−e−z/λ

]λ0

=[1− 1

e

]= 0.632

2b.∫ 2λ

λp(z) dz=

[−e−z/λ

]2λλ

=[1

e− 1

e2

]= 0.233

3.∫ ∞0

p(z) dz=[−e−z/λ

]2λλ

= − [0− 1] = 1.Total probability = 1, since particle must be somewhere!

4. λ =kBT

mg=

1.38.10−23 × 293

4.82.10−26 × 9.81= 8.55 km

5.∫ z∗

0p(z) dz= 0.999 =

[1− e−z∗/λ

]→ e−z∗/λ= 0.001

→ z∗λ

= ln(1000)→ z∗ = ln(1000)λ = 59.1 km

6. P (z) = P0e−z/λ = 1e−z/λ bar, so for P(z) = 0.5 bar

→ e−z/λ = 0.5

→ z

λ= ln 2

z = ln 2× λ = 5.9 km

7. The scale length (8 km) is small compared to the earth’s radius (as is the distance at which99.9% of the gas is contained) since rE ≈ 6400 km

8∗. Ideal gas EOS → PV = NkBT differentiate V dP + P dV = NkBdT

→ −PdV = V dP −NkBdT

But first Law for adiabatic process (dQ = 0); dU = −P dV = nd2NkB dT

where we have used the equipartition of energy for the last bit.

Equating, V dP −NkBdT = nd2NkB dT

V dP =(nd2+ 1

)NkB dT

but γ =nd + 2

nd→ nd =

2

γ − 1→

(nd2

+ 1)=

γ

γ − 1

So → dP =

(γ

γ − 1

)NkBV

dT =

(γ

γ − 1

)(P

T

)dT Q.E.D.

9∗. In eqm. dP = −ρg dz = −mng dz = −mg(P

kBT

)dz where we used P = nkBT

Equating to the above;(

γ

γ − 1

)(��P

T

)dT = −mg

kB

(��P

T

)dz

dT = −mgkB

(γ − 1

γ

)dz = αdz

So T (z) = T (0)− αz where α =mg

kB

(γ − 1

γ

)≈ 9.8K km−1

10∗. For 8.8 km; T = 20− 8.8 ∗ 9.8 ≈ −66◦C(Actually due to heat released by water vapour as it condenses, the lapse rate is actuallyabout 2/3 of what we calculated, so the temperature would only be about −40◦C)

From [9∗.];dPP

=

(γ

γ − 1

)(dTT

)

Integrate: lnP =(

γγ−1

)lnT + lnC

→ PTγ

1−γ = C

P = P (0)(T (0)T

) γ1−γ ≈ P (0)

(293206

) 1.40.4 ≈ 0.29 atmospheres.

This compares to P = P (0) exp(−z/λ) = P (0) exp(−8.5/8.8) = 0.355

which is a little less (but isothermal model gives a good estimate).


Classwork III - AnswersDerivation of the Ideal Gas Equation of State

1. Total no. of particles with x-component of velocity between vx and vx + dvx is

dN = Nf(vx)dvxSo the number density of these particles dn = dN/V = (N/V ) f(vx) dvx = nf(vx) dvx

2. In time ∆t particles with a x displacement of vx ·∆tfrom wall can reach it,

volume occupied by such particles which hit an areaL2 = L2 · vx∆t

and so number of such particles,dNcoll = L2vx∆t× nf(vx) dvx

3. We don’t need extra 12

since we are only going to consider particles travelling towards thewall by integrating between 0 and ∞ (i.e. forward going particles), even though vx couldhave any value between +∞ and −∞

4. Rate at which particles with vx → vx + dvx strike the wall per unit area;

R =1

L2

dNcoll

dt≈ dNcoll/L

2∆T = nvxf(vx) dvx

As above, integrating only over forward going particles, vx,

R =

∫ ∞0

Anvxe−αv2x dvx

= nA

[e−αv

2x

−2α

]∞0

=n

−2α

(απ

)1/2[0− 1]

=n

2

(1

πα

)1/2

=n

2

(2kBT

πm

)1/2

=1

4n

(8kBT

πm

)1/2

= 14n〈v〉

5. momentum change of particles with vx at the wall in time ∆t is

∆p = 2mvx × dNcoll = 2mvx × L2vx∆t× nf(vx) dvx

So force on wall due to these particles is

dF = ∆p/∆t = 2mvx L2vx��∆t nf(vx) dvx/��∆t = 2mvx L

2vx nf(vx) dvx

and pressure is given by dP = dF/L2 = 2mvx��L2vx nf(vx) dvx/��L2

→ dP = 2mnv2x f(vx) dvx

6. P =

∫dP =

∫ ∞0

2mnv2x f(vx) dvx (again only forward going particles)

=

∫ ∞0

2mnv2xA e−αv2x dvx

= 2mnA

∫ ∞0

v2x e−αv2x dvx

= 2mnA1

4

( πα3

)1/2= 2mn

(απ

)1/2 1

4

( πα3

)1/2= mn

1

2α= mn

2kBT

2m

→ P = nkBT

7. 14n〈v〉 ∗ 2m〈v〉 = 1

2nm

(8kBT

πm

)=

(4

π

)nkBT 6= P

This does not take into account that only the vx component of momentum changes (which isthe reason that this is overestimated).

But even taking an average momentum change 〈2mvx〉 does not get this right, because itdoes not account for the fact that faster particles (which carry more momentum) collide withthe walls more often than slower ones. Otherwise you might expect the answer of part 6to be just 1

6n〈v〉, which is what you would get if you assumed all particles had the same

velocity 〈v〉 but were directed equally between the 6 walls of the container.


Classwork IV - AnswersSurface Tension

1. F0 = force acting to left on CD = 2× γl NB γl from each surface

Work done = F0 dx = 2γl dx (ignoring 2nd order term such as dF × dx)

dA = increase in surface area= 2× l dx (each surface increases by l dx)

→ increase in energy = work done = γ dA

2. Remember that each atom in a liquid is bound by ∼ 10 other atoms. But atoms at thesurface only have ∼ 5 nearest neighbours, with a corresponding breaking of ∼ 5 bonds. Bystretching the film, we increase the ratio of surface area to volume, so that the number ofatoms near the surface has increased. Hence the work done has gone into breaking the bondsof this increasing number of surface atoms.

3. Surface area at radius r, A = 4πr2 → dA = 4π × 2r dr = 8πr dr

∴ work done dW = γ dA = 8πγr dr

so Fc =dWdr

= 8πγr

4. Total force on surface = (Poutside × 4πr2)− (Pinside × 4πr2) + 8πγr

But total force = 0 → (Pout × r)− (Pin × r) + 2γ = 0

→ Pin = Pout +2γ

r

5. P = P0 + ρgδ (from lecture 10)

At depth δ = 1 m, P = 1.01× 105 + (103 × 9.81× 1) = 1.11× 105 Nm−2

So Pressure required Pin = 1.11× 105 + (2× 0.0728/10−4) = 1.12× 105 Nm−2

6. To form a bubble (boil) the vapour pressure must be greater than the surface tension, butas seen by F ∝ (1/r), the surface tension is greater for a small bubble. Hence to formbubbles in a pure liquid is very difficult, since it requires the coming together of enoughmolecules with sufficient energy to evapourate. This inhibits boiling and allows the liquidto be superheated. If there are impurities (larger bubbles, or nucleation points), this reducesthe surface tension allowing bubbles to form more easily. Once a bubble forms, as it growsthe surface tension reduces, making it easier to grow further, and explaining the ‘explosive’nature of boiling.

Structure of Matter 23.03.11 First Year Physics Classwork ...

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