Thermal Physics (Unit 5) Flashcards Preview

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Flashcards in Thermal Physics (Unit 5) Deck (24)
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1
Q

Definition of Specific heat capacity

A

Energy required to raise the temperature of 1 kg of a substance by 1 Kelvin.

2
Q

Units of specific heat capacity

A

Jkg-1K-1

3
Q

Definition of Specific latent heat of vapourisation

A

Energy required to change state of 1 kg, liquid to gas, without change of temperature.

4
Q

Definition of Specific latent heat of fusion

A

Energy required to change state of 1 kg, solid to liquid, without change of temperature.

5
Q

Units of specific latent heat

A

Jkg-1

6
Q

Definition of absolute zero

A

Temperature at which there is zero kinetic energy per particle

7
Q

Absolute zero in degrees Celcius

A

0 Kelvin = -273 degrees Celcius (-273.15)

8
Q

Converting between Kelvin and Celcius temperature scales

A

Kelvin to Celcius : subtract 273
Celcius to Kelvin : add 273
Remember 0 degrees C = 273 K

9
Q

Graphical variations between pressure, volume and temperature

A

See sheet

10
Q

Why might energy put into a material not equal the specific heat capacity calculated?

A
  • Energy lost to surroundings

* Specific heat capacity may not be constant over the temperature range used

11
Q

Define terms N, m and Crms in pV=⅓Nm(Crms2).

A

N – number of particles(molecules)
m – mass of individual particles (molecules)
Crms – square root of mean square speed

12
Q

Average kinetic energy (per particle) of a gas at temperature T

A

= 3/2kT = 1/2m(Crms2)

13
Q

Total kinetic energy of gas at temperature T

A

= N3/2kT = N1/2m(Crms2)

Where N is the number of particles/molecules in gas

14
Q

Assumptions of an ideal gas

A
  • All molecules/atoms are identical
  • Molecules/atoms are in random motion
  • Gas contains a large number of molecules
  • The volume of gas molecules is negligible (compared to the volume occupied by the gas) or reference to point masses
  • No forces act between molecules except during collisions
  • Collisions are elastic
  • Collisions are of negligible duration (compared to the time between collisions)
15
Q

What is meant by random motion in context of particles?

A
  • Particles have no preferred direction

* With a range of speeds

16
Q

What is meant by all collisions being elastic?

A

A collision in which momentum and kinetic energy are conserved.

17
Q

Why is average velocity of particles in a container zero?

A
  • Velocity is a vector
  • Random velocity means no preferred direction of movement
  • So velocities cancel
18
Q

By considering motion of particles explain how they exert pressure on the wall of a container

A
  • Molecules/particles have momentum
  • Momentum change at wall
  • Momentum change at wall/collision at wall leads to force
19
Q

In terms of kinetic theory, explain why pressure decreases when gas is removed from a container

A
  • Less gas so fewer molecules
  • So change in momentum per second / rate of change of momentum, is less
  • Pressure is proportional to number of molecules (per unit volume)
20
Q

In terms of kinetic theory, what effect does reducing temperature of a gas have on the pressure

A
  • Pressure decreases
  • Since molecular collisions with wall less frequent
  • And rate of change of momentum is less
21
Q

How does increasing temperature affect the motion of gas particles

A
  • Increase in temperature causes an increase in the mean kinetic energy
  • So mean square speed increases
22
Q

Quantities that increase when temperature of a constant mass of gas at a constant volume is increased

A
  • Pressure

* Average kinetic energy per particle

23
Q

Why are mean square speeds of two different gas molecules at the same temperature not the same

A
  • The (mean) kinetic energy is the same for both gases
  • But masses of atoms is different
  • Mean square speed is proportional to 1/mass
  • Lighter mass atom/molecules have to have higher mean square speed in order to have same average kinetic energy.
24
Q

Derivation of ideal gas equation from kinetic theory

A

Particle, of mass m, travelling with velocity v in a box of length a, width b and height c. Travelling back and forwards along length a and hitting end of area bc.
• Change in momentum of a molecule hitting the end wall p = - 2mv
• from Newton’s 2nd Law : F = p/t
• Time between collisions of a molecule hitting the end wall t = 2a / v
• Force, F, on the molecule hitting the end wall F= - mv2 / a
• from Newton’s 3rd Law : force on molecule is equal and opposite to force on box
• Force, F, on wall from one molecule hitting the end wall; F = mv2 / a
• Pressure = Force / Area
• Area of end wall = bc
• Pressure, p, from one molecule hitting end wall; p = mv2 / abc
• Volume V = abc;
• Pressure, p, from one molecule hitting end wall; p = mv2 / V
• Pressure, p, from N molecules hitting end wall; p = N mv2 / V
• Particles have a range of speeds. Average of speeds squared = (crms)2
• Pressure, p, from N molecules with a range of speeds hitting end wall: p = Nm(crms)2 / V
• On average in a box, 1/3 of molecules will move in each of the 3 directions.
• Ideal gas equation pV = 1/3 Nm(crms)2