13 MATTER: VERY SIMPLEThe gas laws

• Deduce the relationships between pressure, volume and temperature of a gas from experimental data

• Combine them to give the Ideal Gas Law

Boyle’s Law

1627-91pV=constant

At a constant temperature the pressure p and volume V of a gas are inversely proportional

What graph could we draw to test this relationship?!?

Pressure and volume of gases increasing with temperature

Pressure and volume extrapolate to zero at same temperature –273.16 CSo define this temperature as zero of Kelvin scale of temperature, symbol T

Constant volume

heat gas:pressureincreases

T/C

pressure p

45.1

Constant pressure

heat gas:volumeincreases

T/C

45.1

–273 0temperature/C

–273 0temperature/C

2730temperature/K

2730temperature/K

pressure proportional to Kelvin temperature volume proportional to Kelvin temperature

Pressure and volume proportional to absolute temperature

pT VT

2

34

1

volume V

Charles’ Law

1780’s

V/T=constant

At constant pressure, the volume V of a gas is directly proportional to its absolute temperature T.

Pressure LawAt constant volume, the pressure p of a gas is directly proportional to its absolute temperature T.

p/T=constant

Absolute TemperatureAbsolute temperature

measures how hot an object is on a scale starting at absolute

zero (lowest possible temperature, when all particles

have minimum possible energy).

Measured in: Kelvin, K Scale: 1K = 1⁰C Conversion: K = C + 273

1800’s

E.g.. 0 ⁰C = 273K, 10 ⁰C = -263K, -273K = 0 ⁰C

Testing Charles’ Law and the pressure Law…

1. Test the laws using the experimental setup suggested on your sheet.

2. Take measurements of volume or pressure at 4/5 different temperatures (doesn’t matter what temps as long as good range)

3. Plot a rough graph of your results.

Don’t forget to use absolute temperature…Conversion: K = C + 273

What is a mole?One mole has 6.023 x 1023

particles in it.It has a mass equal to the atomic mass in grams:1 mole of Carbon = 12gI mole of Hydrogen = 1g

One law for all gases

2

34

1

volume V

Boyle’s law

pressure p

compress gas:pressure p increasesconstant temperature T

2

34

1

number N

Amount law

pressure p

add more molecules:pressure p increasesconstant temperature T

2

34

1

Pressure law

pressure p

heat gas:pressure p increasesconstant volume V

T/K

45.1

volume V

Charles’ law

heat gas:volume V increasesconstant pressure p

T/K

45.1

p1/V

pN

pT

VT

Combine therelationships into one

pN/V

pVNor

pVintroduceconstant k:pVNkT

combine:

combine:

Combine unknown Nand k into measurablequantity R

Number of moleculesN not known

Nk can be measured:Nk = pV/TFor one mole, defineR = NAk

For n moles:pV = nRT

pVT

k = Boltzmann constantNA = Avogadro number(number of molecules per mole)R = molar gas constant = 8.31 J K –1 mol–1

measured from pV/T for one mole

When NA could be measured:Avogadro number NA = 6.02 1023 particles mol–1

R = molar gas constant = NAk = 8.31 J K–1 mol–1

Boltzmann constant k = 1.38 10–23 J K –1 mol–1

combine:

4

50

1

23

105

N m–2

pV = nRT

n is the number of molesR = molar gas constant

(JK-1mol-1)

One beautiful equation….

Energy the gas has per Kelvin per mole

k = R/NA

Boltzmann constant is derived thus:

pV = nNAkT = nRT

JK-1per particle

Energy the gas has per Kelvin per particle

Summary….pV = constant (if T is constant)V/T = constant (if p is constant)p/T = constant (if V is constant)

Combine to give the Ideal Gas law…pV = nRT or pV = NkT

Where…n = number of moles (amount of gas)R = Molar gas constant (8.31J mol-1K-1)k = Boltzmann constant (1.38x10-23JK-1)N = number of particles

Ideal Gases

Develop problem solving skills involving the gas laws.

Derive the relationship density = PM/RT

A sealed aluminium alloy flask contains air at atmospheric pressure and a temperature of 27 oC. The alloy’s melting point is 620 oC, and the flask will burst if the pressure exceeds 2.9 atmospheres.

Will the flask melt before it bursts, or burst before it melts?

Determining the density of airA plastic vessel (40 cm x 40 cm x 40 cm) containing air at atmospheric pressure is weighed. Its mass is recorded as 371.2 g. The air in the vessel is pumped out and the vessel re-sealed. Its mass is now 300.3 g. From these results, work out the density of air in kg m-3. ρ = mass/volume kgm-3

ρ = density

Upthrust…Where does the force come from that causes a hot air balloon to rise?

Archimedes:Any object in a fluid displaces a volume of the fluid that weighs more than the object does, will float upwards.

Bye Bye Kitty………..A sadistic physicist decides to find out how many helium party balloons need to be attached to a kitten to make it take off. He knows that the upthrust force on a helium balloon is given by the weight of air displaced by it. Mass of kitten = 0.25 kgVolume of a party balloon = 0.027 m3

Mass of balloon (excluding gas in it) = 0.005 kgAir density = 1.1 kg m-3

Helium density = 0.15 kg m-3

Work out how many balloons are needed to make kitty fly!

ρ = m/V kgm-3 ρ = density

Hint…………………..

Hint: For one balloon, work out the net upward force by calculating the up thrust force and the weight of balloon plus contents. Hint: For one balloon, work out the net upward force by calculating the up thrust force and the weight of balloon plus contents.

For one balloon, work out the net upward force by calculating the upthrust force and the weight of balloon plus contents.

Derive…

ρ = pM/RT

ρ = densityp = pressureM = molar massR = universal gas constantT = Absolute Temperaturen = number of molesm = mass

Using the following….pV=nRT, ρ=m/V, n=m/M

Gas densityInstead, you are going to derive an equation for gas density in terms of the intensive properties of pressure, molar mass and temperature, all of which are easy to determine.

1. Write down the ideal gas equation (molar form)

2. Write an expression for n, the number of moles in terms of the mass of gas, m, and the molar mass M.

3. What is the standard expression for the density of a substance?

4. Substitute equation 2 into equation 1 and rearrange to give the correct expression for density on one side.

5. Use your new equation to calculate the density of air at atmospheric pressure (105 Pa) and 25 oC.

Take the molar mass of air as 28.6 g mol-1.

Kinetic theory of gases• Derive PV = nRT from a kinetic

theory standpoint

• Derive expressions for kinetic energy and molecular speed from the resulting equations

Kinetic theory of gases

LawRelationship in terms of variables

Kinetic theory explanation

Boyle’s Law P α 1/V

Pressure Law

Charles’ Law

“Amount” Law

Kinetic TheoryA model that attempts to explain the gas laws.First some assumptions (of an ideal gas)• Gas contains large number of particles• Molecules move randomly at random speeds.• All collisions between wall and particles are

elastic. • Low density so space occupied by molecules is

zero• Energy of motion large enough so that attractive

forces can be ignored.

Momentum Recap…

p = mvF= Δp/Δt

Impact of a ball on a wall (Qs 50S) (3mins)A ball of mass 2 kg, moving at 12 ms–1, hits a massive wall head-on and stops.1. How much momentum did the ball have before impact? 2. How much momentum does the ball have when it has stopped, after impact? 3. How much momentum did the ball lose during impact?4. Assume that Newton’s third law is correct and applies to this case. How much momentum did the wall gain?

24 kgms-1

24 kgms-1

24 kgms-1

0 kgms-1

Force due to a stream of elastic ballsSuppose the wall is hit by a stream of 2 kg balls. In this case each ball arrives with a speed of 12 m s–1 and bounces straight back with an equal speed of 12 m s–1 in the opposite direction. As previously, 1000 balls arrive at the wall in 10 s.

8. Calculate the change of momentum when one ball arrives at the wall and bounces away. 9. Calculate the change in momentum for all the 1000 balls.10. Calculate the average force on the wall during that 10 s period.

48 kg m s–1

48000 kg m s–1

4.8 kN

Deriving the ideal gas equation from kinetic theory

Consider a cuboid, dimensions x,y,z, containing a single gas molecule of mass m travelling at speed v along the X direction. The molecule collides elastically with wall YZ andthen travels in the opposite direction towards the other YZ wall.Q1. Show that the time between collisions of the molecule with

the same wall is t = 2x/v.Q2. What is the rate at which the molecule collides with the wall?Q3. The molecule collides elastically with the wall and rebounds

in the opposite direction. What is its change in momentum?Q4. Newton’s second law can be stated as Force = rate of change

of momentum. Use this together with your answers to the previous questions to show that the average force on YZ due to the molecule colliding with it is F = mv2/x .

Q5. Use the basic definition of pressure (pressure = force/area) and your answer to the last question to derive an expression for the pressure P exerted on wall YZ by the single gas molecule.

Kinetic model of a gas

Use change of momentumTo start: one molecule in a box

zx

v y

round trip-time between collisions t = 2x /v

collisions per second = v /2x

end wall of box

momentum +mv before

momentum –mv after

impulse on wall

wall has change inmomentum +2mv

ball has change inmomentum –2mv

momentum 2mv given to wall at each collision

Force = rate of change of momentumforce on wall =momentum per collision collisions per second

force on wall = mv2/x

2mv v /2x

impulse each time molecule returns

force

time

t

zx

y

Calculate pressure = force on wall/area of wall

force on wall = mv2/x

pressure = mv2/xyz

pressure p = mv2/V

(area = yz)

(V = xyz)

force

timeN times as many collisions per second

N moleculespressure p = Nmv2/V

area of wall = yz

xyz = volume V

pressure p

add many molecules all doing the same improve model

force

time1/3 as many collisions per second1/3 of molecules

in each direction,on average

force

time

pressure p = Nmv2/V

average impulse stays the same

take averageover v2

improve model

pressure p = Nmv2/V

allow molecules to move in random directions improve model

allow molecules to move at random speeds

13

13

The kinetic theory of gases predicts that pV = Nmv213

Kinetic model of a gas

Use change of momentumTo start: one molecule in a box

zx

v y

round trip-time between collisions t = 2x /v

collisions per second = v /2x

end wall of box

momentum +mv before

momentum –mv after

impulse on wall

wall has change inmomentum +2mv

ball has change inmomentum –2mv

momentum 2mv given to wall at each collision

Force = rate of change of momentumforce on wall =momentum per collision collisions per second

force on wall = mv2/x

2mv v /2x

impulse each time molecule returns

force

time

t

zx

y

Calculate pressure = force on wall/area of wall

force on wall = mv2/x

pressure = mv2/xyz

pressure p = mv2/V

(area = yz)

(V = xyz)

force

timeN times as many collisions per second

N moleculespressure p = Nmv2/V

area of wall = yz

xyz = volume V

pressure p

add many molecules all doing the same improve model

force

time1/3 as many collisions per second1/3 of molecules

in each direction,on average

force

time

pressure p = Nmv2/V

average impulse stays the same

take averageover v2

improve model

pressure p = Nmv2/V

allow molecules to move in random directions improve model

allow molecules to move at random speeds

13

13

The kinetic theory of gases predicts that pV = Nmv213

Deriving the ideal gas equation from kinetic theoryNow we need to extend the model to a more realistic situation where many gas molecules are colliding with the walls.

Q1. If there are N molecules in the gas, how many, on average, will be travelling in the x direction?

Q2. Using your expression for the pressure due to one molecule and your answer to the previous question, show that the pressure due to all molecules is given by P = Nmv2/3V .

Q3. We normally write the equation from the previous question as PV = 1/3 Nmv2, where we now allow the molecules to travel at a range of speeds characterised by a mean square average speed. Can you see any similarities with the experimental ideal gas law?Q4. Write down an expression for the kinetic energy of a gas of N molecules of mass m.Q5. Combine the expression from the previous 2 questions to show that the kinetic energy of the gas is given by KE = 3/2 PV.Q6. The KE of a substance is proportional to its temperature T. Use this to show that PV α T as in the experimental ideal gas law.Q7. Show that, for molar quantities, KE = 3/2 RT = 1/2 Mv2, where M is molar mass.Q8. Use the equation from the previous question to estimate the speed of nitrogen

molecules in the room around you.

Kinetic model of a gas

Use change of momentumTo start: one molecule in a box

zx

v y

round trip-time between collisions t = 2x /v

collisions per second = v /2x

end wall of box

momentum +mv before

momentum –mv after

impulse on wall

wall has change inmomentum +2mv

ball has change inmomentum –2mv

momentum 2mv given to wall at each collision

Force = rate of change of momentumforce on wall =momentum per collision collisions per second

force on wall = mv2/x

2mv v/2x

impulse each time molecule returns

force

time

t

zx

y

Calculate pressure = force on wall /area of wall

force on wall = mv2/x

pressure = mv2/xyz

pressure p = mv2/V

(area = yz)

(V = xyz)

force

timeN times as many collisions per second

N moleculespressure p = Nmv2/V

area of wall = yz

xyz = volume V

pressure p

add many molecules all doing the same improve model

force

time1/3 as many collisions per second1/3 of molecules

in each direction,on average

force

time

pressure p = Nmv2/V

average impulse stays the same

take averageover v2

improve model

pressure p = Nmv2/V

allow molecules to move in random directions improve model

allow molecules to move at random speeds

13

13

The kinetic theory of gases predicts that pV = Nmv213

p = 1/3 Nmv2 / VFrom the 3

dimensions of the box

Average of square of speeds of molecules

Mass of one molecule

v1 v2

v3v4

v5v6v7

pV = 1/3Nm2

pV = NkT

Compare….&

Average KE of a molecule = 3/2kTTotal KE of N molecules = 3/2NkT

Kinetic energy of one mole of molecules U = 3/2RT

r.m.s speed, (Root Mean Square Speed)

A way of averaging speeds.

The distribution of speeds of

molecules in nitrogen at 300 K is

shown here.

0.10

0.08

0.06

0.04

0.02

0.00

speed in metre per second

most probablespeed

mean speed

root meansquare speed

r.m.s speed, Remember…Average KE of a molecule = 3/2kT

So… 1/2m= 3/2kTAverage of square of speeds of molecules

Therefore…. = And …. vr.m.s = =

Kinetic Theory

Develop problem solving skills associated with kinetic theory of gases, including the calculation

of r.m.s speeds.

(Part 2)

Starter:Estimate the speed of air molecules in this room

Speed of a nitrogen moleculeAssume warm room temperature T = 300 K

mass of 1 mole of N2 = 28 10–3 kg mol–1

Avogadro constant NA = 6 1023 particles mol–1

Boltzmann constant k = 1.38 10–23 J K–1

kinetic energy of a molecule

from dynamics from kinetic model

mv212 kT3

2v2 = 3 kTm

mass m of N2 molecule calculate speed

v = 500 m s–1 approximately

Air molecules (mostly nitrogen) at room temperature go as fast as bullets

m = mass of 1 mole of N2

Avogadro constant NA

m =

m = 4.7 10–26 kg

v2 = 3 1.4 10–23 J K–1 300 K

4.7 10–26 kg

v2 = 2.7 105 J kg–1 [same as (m s–1)2]28 10–3 kg mol–1

6 1023 mol–1

Diffusion of gases and liquids• Observe and explain diffusion effects in liquids and gases

• Derive and apply the Einstein equation for diffusion

Evidence for moving molecules

Bromine expanding into a vacuum

remove air fromtube using vacuumpump

attach brominecapsule insealed tube:break capsule

open tap:bromineinstantly f illswhole tube

to vacuumpump

tap opentap closed

bromine

vacuumair in tube bromine in tube

attach brominecapsule insealed tube:break capsule

tap closed

air in tube

Bromine diffusing into air

bromine

bromine verygraduallydif fuses upthe tube

tap open

bromine diffuses into tube

Diffusion shows that molecules move. Rapid diffusion into a vacuum demonstrates high molecular speeds

Random walk in 1 dimensionSuppose a molecule can either move one step to the right (R) or one step to the left (L) as a result of collisions with neighbouring molecules. There is a 50 % chance of stepping to the right as a result of a collision, and a 50 % chance of moving to the left. Assume the distance moved in a step is always the same. Consider a molecule that starts in a certain place and then experiences 10 collisions.

Q1. How many different outcomes are there for a sequence of 10 steps?(Hint: one such outcome is LRLLLRRLRL)

Q2. What is the maximum distance from the start, measured in number of steps, that a molecule could end up after 10 collisions? How likely is this outcome?

Q3. Consider a molecule that ends up at 8R after 10 steps. How many different ways are there of reaching this position in 10 steps? Enumerate them.

Q4. What about a molecule that ends up at 6R after 10 steps? How many different ways are there of reaching this position in 10 steps?

Q5. Which is the most likely place for a molecule to end up after 10 steps? Explain your answer.

Q5. Can you sketch a graph to show the distribution of molecules about the starting position after 10 steps? Explain the shape of the graph you have drawn.

Random walk distribution after 10 steps

10L 8L 6L 4L 2L 0L 2R 4R 6R 8R 10R0

50

100

150

200

250

300

n

Gaussian or Normal distribution with standard deviation = √N = √10 = 3.2

In general, if N steps are taken, estimated distance diffused D = √N x d, where d is the “step length” or mean free path .

Diffusion of perfume across a roomHow far do molecules of perfume diffuse in 2 seconds?

Take the rms speed of the molecules as 500 ms-1, and the mean free path (step length or average distance between collisions) as 10-7 m.

Hint: Work out the distance gone in 2 seconds, and then work out how many collisions are experienced in travelling this distance.

A sixth form student is still recovering from a big night out and randomly walks around not knowing where he is going.

Sixth form centre

Estimate how many steps will it take for

a student to randomly walk from

common room to this lab

Lab

Diffusion of photons through the radiative zone of the Sun

Photons that are created during fusion in the core of the Sun can take a very long time to reach the surface and be emitted.

This is because they undergo a vast number of collisions as they are scattered by protons, electrons and other particles in the radiative zone.

In this exercise, your task is to estimate how long it takes a photon to traverse the radiative zone as it undergoes many randomising collisions.

Diffusion in the Sun

Data: Depth of radiative zone: 109 mMean free path of photon: 0.01 m (distance between successive collisions)Speed of photon: 3 x 108 ms-1

1 year = 3 x 107 seconds Q1. Using the random walk model, how long does it take a photon to travel through the

radiative zone? Q2. Will more photons take a greater or less time to transit the radiative zone than the

value calculated in Q1? Q3. What factors will affect the mean free path of a photon, and how will they affect it? Q4. Predict and explain the effect of increasing the mean free path on the time taken to

transit the radiative zone.

Internal energy and specific thermal capacity

• Explain and use the equation E = mc

• Explain the consequences of the anomalous STC of water

• Explain the pattern in molar STC values for metals

• Apply the First Law to thermodynamic problems

Starter: What is the kinetic energy of 1 mole of an ideal gas at 300 K? Use R = 8.3 J K-1 mol-1.How much energy would you need to supply to raise the temperature of 1 mole of an ideal gas by 1 degree K?

pV = 1/3Nm2

pV = NkT

Compare….&

Average KE of a molecule = 3/2kTTotal KE of N molecules = 3/2NkT

Kinetic energy of one mole of molecules U = 3/2RT

Molar Specific Thermal CapacityU = 3/2 RT (internal

energy)Molar STC = dU/dT = 3/2

R“The molar specific thermal capacity (C) of a substance is the amount of

energy needed to raise the temperature of 1 mol of substance by

1K (1⁰C)”Unit: J mol-1 K-1

Specific Thermal Capacity“The specific thermal capacity (c) of a

substance is the amount of energy needed to raise the temperature of

1kg of substance by 1K (1⁰C)”Unit: Jkg-1K-1

How much energy to have a bath?What do we need to know?

- Mass of water - STC of water- Temp change

ΔE=mcΔθ(at constant volume only)

STC of metalsMetal Molar mass

(kg mol-1)Specific thermal capacity(J kg-1 K-1)

Specific thermal capacity(J mol-1 K-1)

Aluminium 900Copper 390Lead 130Iron 450Mercury 140Magnesium 1020Silver 230Platinum 130Gold 130

Complete the table and comment on the result.Do you see any connection with the molar STC values for monatomic gases?

E = mcQ1. The STC of water is 4200 J kg-1 K-1, much higher than the STC values

of most other materials (see p111). Convert this STC into molar units and compare it with the molar STC of ethanol, 112 J K-1 mol-1.

Q2. The anomalous STC of water can explain many phenomena. Do the following exercise to predict and explain which way the wind blows at the sea side on a sunny day.

(a) Solar radiation of intensity 1000 W m-2 is incident on a 1 m3 “block” of sea water for 1 hour. Calculate the temperature rise in the sea water, assuming all the energy is absorbed. Density of water = 1000 kg m-3.

(b) Consider a 1 m3 block of land rock. Calculate the corresponding temperature rise in this block of rock, assuming the same intensity and heating time. Density of rock = 2700 kg m-3, STC of rock = 880 J kg-1 K-1.

E = mc(c) If the air above the land and the sea is heated efficiently

by thermal contact with those bodies, can you explain the origin of breezes coming off the sea during the day?

(d) Sail powered fishing boats used to leave harbour at night. Explain why.

(e) Can you explain why, even in the absence of the Gulf Stream, the British Isles would have a more temperate climate than other countries on the same latitude such as Canada and Russia?

STARTERQ1. Here are the specific thermal capacities for some gases at 300 K, in units of J g-1 K-1. Use their molar masses to convert them into units of J mol-1 K-1.Ar 0.31 02 0.66 I2 0.145N2 0.74 CO 0.74

Q2. Can you see any patterns in the results?

Q3. Can you explain the trends/patterns you see? Hint: What is the STC for an ideal gas (with translational KE only)?

The First Law of Thermodynamics

• State and explain the First Law of Thermodynamics

• Use the First Law to solve problems where gases do work

Transfers of energy to molecules in two ways

Hit the molecules yourself

molecules speeded up piston pushed in

work done = force distance

Let other molecules hit them

cool gas or other material hot wall

thermal transfer = mc

when both ways are used:

Engine designers arrange to transfer energy by way of heating and by way of doing work

change in internal energy U

work done W plus thermal transfer Q=

H ere the p is ton strikes m olecu les and g ives extram om entum and so extra k inetic energy.

energy transferred = work W done

H ere the m olecu les in the ho t w all h it o ther m olecu leshard and on average g ive them extra k inetic energy.

energy transferred = energy Q transferred thermally

U = W + Q

First law of ThermodynamicsEnergy cannot be created destroyed. However it can be transferred from

one place to another.

ΔU = W + QΔU – Change in internal energy of a materialW – work done on materialQ – Energy transferred thermally

Transfers of energy to molecules in two ways

Hit the molecules yourself

molecules speeded up piston pushed in

work done = force distance

Let other molecules hit them

cool gas or other material hot wall

thermal transfer = mc

when both ways are used:

Energy can be given to molecules by heating and by doing work

change in internal energy U

work done W plus thermal transfer Q=

H e re th e pis to n s trikes m ole cu le s an d g ive s e x tram o m e ntum a n d so ex t ra k in et ic en e rgy.

energy transferred = work W done

He re th e m o le cu le s in th e ho t wa ll hit oth e r m o le cu le sh ard a nd o n a vera g e give th em ext ra k in e tic en e rgy.

energy transferred = energy Q transfer red thermally

U = W + Q

The First Law of ThermodynamicsIn what follows, we will consider ideal gases, but the First Law applies to all substances.

Q1. Energy can be supplied to a gas in two ways. Describe them.

Q2. Write down the equation that summarises the First Law, identifying the terms clearly. What property of the gas gives an indication of its internal energy?

Q3. If a gas expands and does work against its surroundings, and no heat energy is lost or supplied, does the gas lose or gain internal energy?

Q4. If heat energy is lost from a gas and it does no work, what can you say about its final internal energy?

Q5. For the following changes, predict and explain whether the final temperature of the gas will be higher or lower than at the start:

(a) 2000 J of work are done on the gas and 3000 J of thermal energy are supplied.(b) The gas does 2000 J of work and 1500 J of thermal energy are supplied to it.(c) 4000 J of work are done on the gas and 3000 J of thermal energy are supplied to it.

Q6. (a) What is the internal energy of 100 moles of argon gas at 25 oC? (R=8.3 JK-1mol-1) (b) What volume would the gas occupy at 100 000 Pa pressure? (c) The gas is allowed to push back a piston of area 0.25 m2 over a distance of 15 cm. Calculate the work done by the

gas. (Hint: Work = force x distance. Ignore any pressure change.) (d) If 4000 J of heat energy are lost by the gas during the expansion, calculate its final internal energy and hence its final

temperature.

The first law

An ideal gas at 27 oC is contained in a cubic container of side length 50 cm. The gas pressure is 1x 105 Pa.

Q1. Calculate the number of molecules of gas in the container. (k=1.4 x 10-23 J K-1)

Q2. Estimate the internal energy of the gas in the container.

Q3. The container is compressed on all sides until each side has shortened in length by 10 %. Calculate the new pressure, assuming that the compression happens isothermally (no change in temperature).

Q4. Compressing the gas involves doing work. Where does this energy go if the compression of the gas is being carried out isothermally?

Q5. Estimate the work done on the gas when it is compressed. Hint: Work = force x distance = pressure x volume change

Q6. If the container is instead thermally isolated when the compression takes place, estimate the final temperature of the gas.

Otto Engine...

Determining the temperature of a Bunsen flame

• Devise and carry out an experiment to determine the temperature of a roaring Bunsen flame

• Consider critically random measurement uncertainties and systematic errors

Starter: A blacksmith heats a 1.1 kg iron horseshoe in a forge. When the shoe is very hot, he plunges it into a tub of 15 kg of cold water at 10 oC. The temperature of the water is observed to rise by 10 oC. To what temperature had the horseshoe been heated?STC of water= 4200 J kg-1 K-1 ΔE = m c Δ ΘSTC of iron = 450 J kg-1 K-1 What assumptions do you need to make?

ΔE = m c Δ ΘA blacksmith heats a 1.1 kg iron horseshoe in a forge. When the shoe is very hot, he plunges it into a tub of 15 kg of cold water at 10 oC. The temperature of the water is observed to rise by 10 oC. To what temperature had the horseshoe been heated?

STC of water= 4200 J kg-1 K-1 STC of iron = 450 J kg-1 K-1

Uncertainty and Error

Random uncertainties• Identify sources of

measurement uncertainty and quantify them.

• Which contributes the most to the uncertainty in the final answer?

Systematic errors• Identify limitations in

the design of the experiment and explain how they could either be eliminated or reduced in effect.