Ideal gases

Question 1

Two factors that affect the force, and hence the pressure, that the gas exerts on the box:

Answer 1

• the number of molecules that hit each side of the box in one second
• the force with which a molecule collides with the wall

Question 2

Effect on pressure with increasing speed of colliding molecules in a container

Answer 2

The higher the speed of the molecule the greater the force that it exerts as it collides with the wall. Hence, the pressure on the wall will increase if the molecules move faster.

Question 3

What happens when the piston in a bicycle pump is pushed inwards, but the temperature is kept constant?

Answer 3

More molecules will hit the piston in each second, but each collision will produce the same force because the temperature and therefore the average speed of the molecules is the same.

Question 4

mass of one unified atomic mass unit, 1u

Answer 4

1 u = 1.66 × 10^−27 Kg

Question 5

No. of atoms or molecules formula

Answer 5

N = n × N avd

Question 6

Boyle’s law

Answer 6

The pressure exerted by a fixed mass of gas is inversely proportional to its volume, provided the temperature of the gas remains constant.
Note that this law relates two variables, pressure and volume, and it requires that the other two, mass and temperature, remain constant.

Question 7

Example with explanation of Boyle’s law

Answer 7

If a gas is compressed at constant temperature, its pressure increases and its volume decreases. A decrease in volume occupied by the gas means that there are more particles per unit volume and more collisions per second of the particles on unit area of the wall. Because the temperature is constant, the average speed of the molecules does not change. This means that each collision with the wall involves the same change in momentum, but with more collisions per second on unit area of the wall there is a greater rate of change of momentum and, therefore, a larger pressure on the wall.

Question 8

p1V1 = p2V2

Answer 8

p1 and V1 represent the pressure and volume of the gas before a change, and p2 and V2 represent the pressure and volume of the gas after the change

Question 9

Absolute zero, 0K

Answer 9

A temperature at which the volume of a gas does, in principle, shrink to zero.

Question 10

Ideal gas equation

Answer 10

pV = nRT
or
pV = NkT
(R = 8.31 Jmol -1 K -1)
(k = 1.38 × 10^-23 J K -1 )

Question 11

Kinetic theory of gases

Answer 11

A theory that links these microscopic properties of particles /atoms /molecules (mass and velocity) to the macroscopic properties of a gas (volume and pressure).

Question 12

Kinetic theory Assumption 1:
A gas contains a large number of particles (atoms or molecules) moving in random directions and collide elastically with the walls and with each other.

Answer 12

Explanation :
Kinetic energy cannot be lost. The internal energy of the gas is the total kinetic energy of the particles.

Question 13

Kinetic theory Assumption 2 :
The forces between particles are negligible, except during collisions.

Answer 13

Explanation :
If the particles attracted each other strongly over long distances, they would all tend to clump together in the middle of the container.

Question 13

Kinetic theory Assumption 3 :
The volume of the particles is negligible compared to the volume occupied by the gas.

Answer 13

Explanation :
When a liquid boils to become a gas, its particles become much farther apart.

Question 14

Kinetic theory Assumption 4 :
The time of collision by a particle with the container walls is negligible compared with the time between collisions.

Answer 14

Explanation :
The molecules can be considered to be hard
spheres.

Question 15

1/3 < c2 > explanation

Answer 15

If there are components cx, cy and cz of the velocity in the x-, y- and z- directions then c^2= cx^2 + cy^2 +cz^2. There is nothing special about any particular direction and so,< cx2 >=< cy2 > = < cz2 > and < cx2 > = 1/3 < c2 >

Question 16

Pressure of an ideal gas Equation

Answer 16

p = 1/3 Nm/V < c2 > = 1/3 Nρ/V < c2 >
and
pV = 1/3 Nm < c2 >

Question 17

Equation to tell us how the absolute temperature of a gas (a macroscopic property) is related to the mass and speed of its molecules

Answer 17

1/3 Nm < c2 > = nRT ( = NKT )

Question 18

kinetic energy of a molecule

Answer 18

3/2 KT = 1/2 m< c2 >

Question 19

Boltzmann constant, k

Answer 19

R / Na = 1.38 × 10 ^-23 JK-1

Question 20

What is translational kinetic energy and what does it depend on?

Answer 20

The energy that the molecule has because it is moving along; a molecule made of two or more atoms may also spin or tumble around, and is then said to have rotational kinetic energy.
The mean translational kinetic energy of an atom (or molecule) of an ideal gas is proportional to the thermodynamic temperature.

Question 21

Two ways to find mean or average kinetic energy

Answer 21

1. Add up all the kinetic energies of the individual molecules of the gas and then calculate the average k.e. per molecule.
2. Alternatively, watch an individual molecule over a period of time as it moves about, colliding with other molecules and the walls of the container and calculate its average k.e over this time.

Question 22

What does Boltzmann constant relate?

Answer 22

It tells us how a property of microscopic particles (the kinetic energy of gas molecules) is related to a macroscopic property of the gas (its absolute temperature). That is why its units are joules per kelvin (JK−1).

Question 23

Define Root mean square speed & Formula

Answer 23

the square root of the average of the squares of the speeds of all the molecules in a gas.
V r.m.s = 3RT/Mr
V r.m.s = √3KT/m

Question 24

Effect of k.e being proportional to T on mean speed

Answer 24

Since mean k.e. ∝ T, it follows that if we double the thermodynamic temperature of an ideal gas (for example, from 300 K to 600 K), we double the mean k.e. of its molecules. It doesn’t follow that we have doubled their speed; because k.e. ∝ ν2, their mean speed has increased by a factor of √2.