ch20 - ideal gases

Question 1

gas particles

Answer 1

. random directions
. straight line motion

Question 2

average speed of molecules in air

Answer 2

400 ms-1

Question 3

atmospheric pressure

Answer 3

100000 Pa

Question 4

surface area of typical person

Answer 4

2.0 m²

Question 5

factors affecting pressure exerted by gas on box

Answer 5

. number of molecules that hit each side of box per second
. force with which a moldcule collides with the wall

Question 6

more pressure

Answer 6

. more molecules
. more temperature

Question 7

properties of gas

Answer 7

. measured in g or kg
. sometimes moles (SI of substance) of gas is considered

1u = 1.66*10^-27 kg
mass/ u * no. = 6.20 @

Question 8

pressure

Answer 8

. normal force/ unit area
. 1 Pa = 1 Nm-2

Question 9

temperature

Answer 9

. celcius, kelvin, fahrenheit
. Tc - 0 / 100 = Tf - 32 / 180 = Tk - 273.15 / 100
. Tk = Tc + 273.15

Question 10

volume

Answer 10

. measure of space occupied by the gas
. m³

Question 11

mass

Answer 11

. measured in g or kg
. sometimes moles (SI of substance) of gas is considered

1u = 1.66*10^-27 kg
mass/ u * no. = avocado’s constant

Question 12

avogadro constant, Na

Answer 12

.the number of particles in one mole of any substance
Na = 6.02 * 10^23 mol-1

Question 13

Boyle’s Law

Answer 13

the pressure exerted by a fixed mass of a gas inversely proportional to its volume, provided the temperature of the gas remains constant

P = k/V
p1V1 = p2V2

Question 14

Charles’ Law

Answer 14

the volume ocupied by a gas at constant pressure is directly proportional to its thermodynamic (absolute) temperature

V = kT
pV/T = constant

Question 15

ideal gas

Answer 15

gas that behaves according to PV=nRT

Question 16

gases approach ideal behaviour at

Answer 16

. high temperature
. low pressure

Question 17

ideal gas equation/equation of state

Answer 17

pV=nRT
pV=NkT

. n is no. moles
. N is no. molecules
. k is Boltzmann constant
. R is gas constant (8.31)

Question 18

number moles

Answer 18

mass/ molar mass

Question 19

kinetic theory of gases

Answer 19

a model based on the microscopic motion of atoms or molecules of a gas

Question 20

assumptions for kinetic theory of gases

Answer 20

. gas contains many particles colliding and moving randomly
. forces between particles are negligible
. volume of particles is negligible compared to volume occupied by gas
. time during collisions is negligible compared to the time between collisions

Question 21

force

Answer 21

F = <>mv/T

Question 22

pressure

Answer 22

F/A

Question 23

derive pV = 1/3 * Nm<c²>

Answer 23

change in p = -mc - (mc)
change in p = - 2mc

time between collisions = d/s
time between collisions = 2l/c

force = change in p/time
force = 2mc/ (2l/c)
force = mc²/l

pressure = F/A
pressure = (mc²/l) / l²
pressure of 1 molecule = mc²/l³

pressure of N molecules in single direction = Nm <c²> / l³

three dimensions:
<cx²> = <cy²> = <cz²>
1/3 <c²>

so pressure = 1/3 * (Nm<c²> / l³)

l³ = volume

so pV = 1/3 * Nm<c²> t

Question 24

KINETIC ENERGY OF MOLECULE = 3/2 * kT derive

Answer 24

pV = 1/3 * Nm<c²>
pV = nRT

1/3 * Nm<c²> = nRT
m<c²> = 3nRT / N
m<c²> = 3RT/Na
1/2 m<c²> = 3RT/2Na

R/ Na = Boltzmann constant, k

1/2 m<c²> = 3/2*kT

KINETIC ENERGY OF MOLECULE = 3/2 * kT

Question 26

Na

Answer 26

N/n

Question 27

k (boltzmann constant)

Answer 27

R/Na units = J K-1 value = 1.38 * 10^-23 JK-1

Question 28

relation between mean translational kinetic energy of atom and time

Answer 28

directly proportional

Question 29

types of kinetic energy

Answer 29

. translational . rotational

Question 30

when is mean speed = root mean square speed

Answer 30

when all molecules have same speed

Question 31

root mean square speed

Answer 31

Crms = () ^1/2 . the square root of thr average of the square of the speeds of all the molecules in a gas