Kinetic Theory of Gases

Physics 202Professor Lee

CarknerLecture 15

Through which material will there be the most heat transfer via conduction?

a) solid ironb) woodc) liquid waterd) aire) vacuum

Through which 2 materials will there be the most heat transfer via radiation?

a) solid iron and woodb) wood and liquid waterc) liquid water and aird) vacuum and solid irone) vacuum and air

Through which 2 materials will there be the most heat transfer via convection?

a) solid iron and woodb) wood and liquid waterc) liquid water and aird) vacuum and solid irone) vacuum and air

PAL #14 Heat Transfer Conduction and radiation through a window

H = kA(TH-TC)/L = (1)(1X1.5)(20-10)/0.0075 =2000 J/s P = A(Tenv

4-T4) P=(5.6703X10-8)(1)(1X1.5)(2934-2834) = 81.3 J/s

Double pane window H = A(TH-TC)/[(L1/k1)+(L2/k2)+(L3/k3)] H =

(1X1.5)(20-10)/[(.0025/1)+(0.0025/0.026)+(0.0025/1)]

H = 148.3 J/s Conduction dominates over radiation and

double-pane windows are about 13 times better

Chapter 18, Problem 39 Two 50g ice cubes at -15C dropped into

200g of water at 25 C. Assume no ice melts

miciTi + mwcwTw = 0

(0.1)(2220)(Tf-(-15) + (0.2)(4190)(Tf-25) = 0

222Tf + 3330 + 838Tf – 20950 = 0

Tf = 16.6 C Can’t be true (can’t have ice at 16.6 C)

Try different assumption

Chapter 18, Problem 39 (cont)

Assume some (but not all) ice melts Tf = 0 C, solve for mi

miciTi + miLi + mwcwTw = 0 (0.1)(2220)(0-(-15) + mi(333000) + (0.2)(4190)

(0-25) = 0 0 + 3330 + 333000mi + 0 – 20950 = 0 mi = 0.053 kg This works, so final temp is zero and not all ice

melts If we got a number larger than 0.1 kg we would know

that all the ice melted and we could try again and solve for Tf assuming all ice melts and then warms up to Tf

Chapter 18, Problem 39 (cont)

Use one 50g ice cube instead of two We know that it will all melt and warm up

miciTi + miLi + micwTiw + mwcwTw = 0 (0.05)(2220)(0-(-15) + (0.05)(333000)

+ (0.05)(4190)(Tf-0) + (0.2)(4190)(Tf-25) = 0

0 + 1665 + 16650 + 209.5Tf - 0 – +838Tf -20950 = 0

Tf = 2.5 C

What is a Gas?

A gas is made up of molecules (or atoms) The temperature is a measure of the random motions of

the molecules

We need to know something about the microscopic properties of a gas to understand its behavior

Mole

1 mol = 6.02 X 1023 units 6.02 x 1023 is called Avogadro’s number (NA)

M = mNA Where m is the mass per molecule or atom

Gases with heavier atoms have larger molar masses

Ideal Gas

Specifically 1 mole of any gas held at constant temperature and constant volume will have the almost the same pressure

Gases that obey this relation are called ideal gases A fairly good approximation to real gases

Ideal Gas Law

The temperature, pressure and volume of an ideal gas is given by:

pV = nRT Where:

R is the gas constant 8.31 J/mol K

Work and the Ideal Gas Law

We can use the ideal gas law to solve this equation

p=nRT (1/V)

Vf

VipdVW

Vf

VidVV1

nRTW

Isothermal Process

If we hold the temperature constant in the work equation:

W = nRT(1/V)dV = nRT(lnVf-lnVi)

W = nRT ln(Vf/Vi)

Isotherms

PV = nRTT = PV/nR

For an isothermal process temperature is constant so:

If P goes up, V must go down

Can plot this on a PV diagram as isotherms One distinct line for each

temperature

Constant Volume or Pressure

W=0

W = pdV = p(Vf-Vi)W = pV

For situations where T, V or P are not constant, we must solve the integral The above equations are not universal

Random Gas Motions

Gas Speed

The molecules bounce around inside a box and exert a pressure on the walls via collisions

How are p, v and V related?

The rate of momentum transfer depends on volume

The final result is:p = (nMv2

rms)/(3V) Where M is the molar mass (mass contained in 1

mole)

RMS Speed

There is a range of velocities given by the Maxwellian velocity distribution

It is the square root of the sum of the squares of all of the velocities

vrms = (3RT/M)½

For a given type of gas, velocity depends only on

temperature

Translational Kinetic Energy

Using the rms speed yields:Kave = ½mvrms

2

Kave = (3/2)kT

Where k = (R/NA) = 1.38 X 10-23 J/K and is called the Boltzmann constant

Temperature is a measure of the average kinetic energy of a gas

Maxwell’sDistribution

Maxwellian Distribution and the Sun

The vrms of protons is not large enough

for them to combine in hydrogen fusion

There are enough protons in the high-speed tail of the distribution for fusion to occur

Next Time

Read: 19.8-19.11 Homework: Ch 19, P: 12, 31, 49,

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