ch14 Flashcards

1
Q

describe the motion of gas molecules

A

gas molecules move with constant brownian motion which is random motion of molecules caused by collisions with larger particles

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2
Q

how do gases exert a force (pressure) on it a container

A

the molecules continuously collide with each other and the walls of the container.

The collisions cause a change in momentum (impulse) which produces a force equal to the rate of change of momentum

this force gives pressure (P = F/A)

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3
Q

how do you get impulse from a force-time graph

A

impulse is equal to the area under a force-time graph

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4
Q

describe the pathing of particles

A

particles take a random path
they don’t travel in a straight line, but are constantly changing direction due to collisions, which is why diffusion is slow

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5
Q

equation for average distance moved (displacement)

A

distance moved = √N * step length (mean free step)
N = number of steps moved
step length = length of one step

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6
Q

boyle’s law

A

when gas is at a constant temperature, pressure P and volume V are inversely proportional
pV = constant

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7
Q

why does boyle’s law still work

A

as volume decreases, particles are closer together, increasing density and colliding more frequently
this exerts a greater force and therefore greater pressure on the container

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8
Q

1 bar = normal atmospheric pressure
the pressure in a car tyre is 2.6 bar
the tyre volume is 6 litres and the temp is constant

calculate the volume of gas that escapes if the tyre is punctured, assuming the tyre keeps its shape

A

pV = constant
first calculate constant
2.6 * 6 = 15.6 bar Litre

air will escape until the tyre pressure matches atmospheric pressure, 1 bar so work out volume after it’s expanded to that pressure
pV = 15.6 bar litre

V = 15.6 / P = 15.6 / 1
volume = 15.6 litres
but 6 litres remains constant in deflated tyre

volume escaping = 15.6 litre - 6 litre = 9.6 litre

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9
Q

how to calculate number of particles

A

number or particles = number of moles * Avogadro constant Na

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10
Q

how is pressure affected when the volume of a gas increases

A

when the volume of a gas increases, the space between molecules increases and so the time between collisions is larger

this cause the rate of collisions and so rate of change of momentum to decrease

this means the force exerted is lower, causing a decrease in pressure

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11
Q

what does Charles’ Law

A

at a constant pressure, volume V is directly proportional to absolute temperature T

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12
Q

number of moles =

A

number of moles =
mass of sample / mass of 1 mole

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13
Q

explain charles’ law

A

as temp increases, the average kinetic energy of the molecules increases

pressure is constant so the force and rate of change of momentum is constant

to keep it like this, the volume increases so the faster speed of molecules is compensated by the larger gaps between them

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14
Q

state pressure law

A

when a gas has a fixed volume, pressure is directly proportional to the absolute temperature

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15
Q

explain pressure law

A

as temperature increases, the average kinetic energy increases so speed of molecules increases

this increases the rate of collisions so producing a larger rate of change of momentum

this leads to a greater force exerted and so an increase in pressure

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16
Q

what does kinetic theory assume gases to be

A

ideal gases

17
Q

assumptions of ideal gases

A

the gas contains a large number of molecules

the gas molecules are identical to each other

all collisions between molecules and container are perfectly elastic (energy is conserved)

time taken for collisions is negligible compared to time between collisions

there are no intermolecular forces so molecules do not attract each other

molecules are in constant random Brownian motion

gas particles obey Newton’s laws of motion

18
Q

ideal gas equations:

A

pV = nRT

p = pressure (Pa)
V = volume (m³)
n = number of moles
R = gas constant 8.314 J mol^-1 K^-1
T = temp in K

pV = NKT

N = number of molecules
K = boltzmann constant
1.38 * 10^-23 J K^-1

19
Q

Boyle’s law relationship

A

pV = constant

20
Q

charles’ Law relationship

A

V / T = constant

21
Q

pressure law relationship

A

p / T = constant

22
Q

how can you work out nR using Nk

A

Nk = nR
N = number of molecules
k = boltzmann constant
n = number of moles
R = gas constant

23
Q

equation for force F caused by a single molecule on a wall

A

F = (mc²) / x
F = force in newtons
m = mass of molecule
c = velocity
x = length of box

24
Q

final theoretical ideal gas equation

A

pV = 1/3 Nmc²

25
Q

how do you get R.M.S, root mean square speed,
the average speed of molecules

A

you find the mean of the speeds squared
c² (small line over c)

26
Q

derivation of 1/2mc(bar)² = 3/2kT

what does the equation tell us

A

we have 2 equations:
pV = 1/3 Nmc² and pV = NKT

mc² is very similar Ek = 1/2mv²
so pV = 1/3 Nmc² =
1/3N(1/2mc²) = but 1/3 * 1/2 doesn’t equal to 1/3, so we must have 2/3 and 1/2 to get the original 1/3

pV = 2/3N(1/2mc²)
NKT = 2/3N(1/2mc²)
KT = 2/3(1/2mc²) so:

3/2KT = 1/2mc²
for a mixture of gases like air the mean kinetic energy is the same at a given temperature

27
Q

approximation of kinetic energy

A

3/2KT

3RT / 2Na

28
Q

what is internal energy and how do you calculate it

A

internal energy is the amount of energy contained in a system

internal energy is the sum of kinetic and potential energies of its particles

29
Q

internal energy in an ideal gas

A

there are assumed to be no intermolecular forces, so no potential energy
so assumption is internal energy = kinetic energy

30
Q

equation of internal energy.
the first law of thermodynamics

A

ΔU = W + Q
ΔU = change in internal energy
Q = energy transferred thermally
W = work done (energy)
you calculate work done from W = force (N) * distance (m)
W and Q can be negative

31
Q

define specific heat (thermal) capacity

A

the energy needed to raise the temperature of a 1kg object by 1 degrees

32
Q

equation for specific heat capacity

A

E (energy transferred) = ΔU (change in internal energy)

E = m c Δθ

E = energy (J)
m = mass (KG)
c = specific heat capacity
θ = temperature (t)

33
Q

equation for power, work done and time

A

power is the rate at which work is done

P = ΔU * t

P = power (W)
ΔU = energy transferred
t = time