Thermal Physics (Unit 5) Flashcards
Definition of Specific heat capacity
Energy required to raise the temperature of 1 kg of a substance by 1 Kelvin.
Units of specific heat capacity
Jkg-1K-1
Definition of Specific latent heat of vapourisation
Energy required to change state of 1 kg, liquid to gas, without change of temperature.
Definition of Specific latent heat of fusion
Energy required to change state of 1 kg, solid to liquid, without change of temperature.
Units of specific latent heat
Jkg-1
Definition of absolute zero
Temperature at which there is zero kinetic energy per particle
Absolute zero in degrees Celcius
0 Kelvin = -273 degrees Celcius (-273.15)
Converting between Kelvin and Celcius temperature scales
Kelvin to Celcius : subtract 273
Celcius to Kelvin : add 273
Remember 0 degrees C = 273 K
Graphical variations between pressure, volume and temperature
See sheet
Why might energy put into a material not equal the specific heat capacity calculated?
- Energy lost to surroundings
* Specific heat capacity may not be constant over the temperature range used
Define terms N, m and Crms in pV=⅓Nm(Crms2).
N – number of particles(molecules)
m – mass of individual particles (molecules)
Crms – square root of mean square speed
Average kinetic energy (per particle) of a gas at temperature T
= 3/2kT = 1/2m(Crms2)
Total kinetic energy of gas at temperature T
= N3/2kT = N1/2m(Crms2)
Where N is the number of particles/molecules in gas
Assumptions of an ideal gas
- 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)
What is meant by random motion in context of particles?
- Particles have no preferred direction
* With a range of speeds
What is meant by all collisions being elastic?
A collision in which momentum and kinetic energy are conserved.
Why is average velocity of particles in a container zero?
- Velocity is a vector
- Random velocity means no preferred direction of movement
- So velocities cancel
By considering motion of particles explain how they exert pressure on the wall of a container
- Molecules/particles have momentum
- Momentum change at wall
- Momentum change at wall/collision at wall leads to force
In terms of kinetic theory, explain why pressure decreases when gas is removed from a container
- 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)
In terms of kinetic theory, what effect does reducing temperature of a gas have on the pressure
- Pressure decreases
- Since molecular collisions with wall less frequent
- And rate of change of momentum is less
How does increasing temperature affect the motion of gas particles
- Increase in temperature causes an increase in the mean kinetic energy
- So mean square speed increases
Quantities that increase when temperature of a constant mass of gas at a constant volume is increased
- Pressure
* Average kinetic energy per particle
Why are mean square speeds of two different gas molecules at the same temperature not the same
- 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.
Derivation of ideal gas equation from kinetic theory
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
