6 Temperature and Ideal Gas

Question 1

Temperature

Answer 1

Measure of degree of hotness of an object

Question 2

Zeroth Law of Thermodynamics

Answer 2

If two bodies are separately in thermal equilibrium with a third body, then these two bodies are in thermal equilibrium with each other

Question 3

Absolute zero

Answer 3

0K/-273.15°C

Question 4

Conversion of Kelvin to Celsius

Answer 4

T/K = T/°C +273.15

Question 5

Triple point of water

Answer 5

273.16 K

Question 6

Ideal gas

Answer 6

A hypothetical gas that obeys the equation of state of an ideal gas perfectly at all pressures, temperatures and volumes

Question 7

Boyle’s Law

Answer 7

When temperature is constant, pV is constant

Question 8

Charles’ Law

Answer 8

When pressure is constant, V/T is constant

Question 9

Gay-Lussac’s Law

Answer 9

When volume is constant p/T = constant

Question 10

Equation of State of an Ideal Gas

Answer 10

pV=nRT, n is amount of gas and R = 8.31 J K^-1 mol^-1
pV=NkT, N is the number of gas molecules and k = 1.38*10^-23 J K^-1

Question 11

One mole

Answer 11

The amount of substance that contains as many elementary particles as there are atoms in 0.012 kg of carbon-12

Question 12

Avogadro’s constant

Answer 12

Number of atoms in 0.012 kg of carbon-12
6.02*10^23 mol^-1

Question 13

Assumptions of kinetic theory of gases

Answer 13

1. Any gas consists of a very large number of molecules
2. Molecules are in rapid and random motion
3. No intermolecular attractive forces
4. Collisions between gas molecules and between gas molecules and the container walls are perfectly elastics
5. Duration of collisions is negligible compared with the time interval between collisions
6. The volume of gas molecules is negligible compared with the volume of the container

Question 14

Root-mean-square speed

Answer 14

c(rms) = √<c^2>

Question 15

Kinetic Theory Equations

Answer 15

p = (1/3) (Nm/V) <c^2>

Question 16

Derivation of Kinetic Theory Equations

Answer 16

1. Elastic collision with a wall by the molecule results in a change of momentum 2 mc(x)
2. Molecule travels a distance 2l before it collides with wall again -> Δt = 2l/c(x)
3. Rate of change of momentum of this molecule due to collision with wall -> Δp/Δt = (1.)/(2.) = mc(x)^2/l
4. Since p=F/A, p = (3.)/l^2 = mc(x)^2/l^3 = m(c)^2/V
5. <c(x)^2> represent the mean value of the sum of the squared velocity components in the x-direction
6. Therefore, p=Nm<c(x)^2> /V
7. To include y and z components, c^2 = c(x)^2 + c(y)^2 +c(z)^2
8. Hence, <c^2> = <c(x)^2> + <c(y)^2> + <c(z)^2>
9. Since number of molecules is large, and are in constant, random motion, <c(x)^2> ≈ <c(y)^2> ≈ <c(z)^2>
   Hence <c(x)^2> ≈ 1/3 <c^2>
10. Thus, from 6., p = (1/3) (Nm/V) <c^2>

Question 17

Mean kinetic energy equation

Answer 17

1/2 m<c^2> = 3/2 kT
Were k = 1.38*10-23 J K-1