Thermal Physics (Unit 5)

Question 1

Definition of Specific heat capacity

Answer 1

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

Question 2

Units of specific heat capacity

Answer 2

Jkg-1K-1

Question 3

Definition of Specific latent heat of vapourisation

Answer 3

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

Question 4

Definition of Specific latent heat of fusion

Answer 4

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

Question 5

Units of specific latent heat

Answer 5

Jkg-1

Question 6

Definition of absolute zero

Answer 6

Temperature at which there is zero kinetic energy per particle

Question 7

Absolute zero in degrees Celcius

Answer 7

0 Kelvin = -273 degrees Celcius (-273.15)

Question 8

Converting between Kelvin and Celcius temperature scales

Answer 8

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

Question 9

Graphical variations between pressure, volume and temperature

Answer 9

See sheet

Question 10

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

Answer 10

• Energy lost to surroundings

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

Question 11

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

Answer 11

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

Question 12

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

Answer 12

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

Question 13

Total kinetic energy of gas at temperature T

Answer 13

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

Where N is the number of particles/molecules in gas

Question 14

Assumptions of an ideal gas

Answer 14

• 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)

Question 15

What is meant by random motion in context of particles?

Answer 15

• Particles have no preferred direction

* With a range of speeds

Question 16

What is meant by all collisions being elastic?

Answer 16

A collision in which momentum and kinetic energy are conserved.

Question 17

Why is average velocity of particles in a container zero?

Answer 17

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

Question 18

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

Answer 18

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

Question 19

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

Answer 19

• 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)

Question 20

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

Answer 20

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

Question 21

How does increasing temperature affect the motion of gas particles

Answer 21

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

Question 22

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

Answer 22

• Pressure

* Average kinetic energy per particle

Question 23

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

Answer 23

• 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.

Question 24

Derivation of ideal gas equation from kinetic theory

Answer 24

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