• Physics 9702 Doubts | Help Page 174

Question 867: [Matter > Gases > Kinetic theory of gases]

Which statement about molecules in a gas is correct?

A In Brownian motion experiments, the molecules can be seen moving randomly in all directions.

B The pressure exerted by a gas is caused by molecules bouncing against each other and changing kinetic energy.

C The pressure exerted by a gas is caused by molecules rebounding from the walls of a container and changing momentum.

D When the average speed of the molecules in a closed container increases, the density must also increase.

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q18

Solution 867:

Answer: C.

Molecules are too small to be seen, even by a microscope. In Brownian motion, the motion of molecules is inferred by the random motion of smoke particles (which are visible). It is suggested that the molecules, which are in random motion, collide with the smoke particles which then move in random motion. [A is incorrect]

Pressure = Force / Area

The force is due to the changing momentum of the molecules and the area is the wall of the container being hit by the molecules. [B is incorrect and C is correct]

The average speed / kinetic energy is relate to the temperature, not density. [D is incorrect]

Question 868: [Ideal gas]

Constant mass of an ideal gas has volume of 3.49 × 10³cm³ at a temperature of 21.0°C. When the gas is heated, 565 J of thermal energy causes it to expand to a volume of 3.87 × 10³ cm³ at 53.0 °C. This is illustrated in Fig.1.

(a) Show that initial and final pressures of the gas are equal

(b) Pressure of the gas is 4.20 × 10⁵ Pa.

For this heating of the gas,

(i) calculate work done by the gas,

(ii) use the first law of thermodynamics and answer in (i) to determine the change in internal energy of gas

(c) Explain why the change in kinetic energy of molecules of this ideal gas is equal to the change in internal energy

Reference: Past Exam Paper – June 2014 Paper 42 Q2

Solution 868:

(a)

Use of temperatures in kelvin

21.0°C = (273+21 =) 294K     and 53.0°C = (273+53 =) 326K

Initially, V / T = (3.49×10³) / 294 = 11.87

For final state of gas, V / T = (3.87×10³) / 326 = 11.87

{Other forms of the ideal gas equation could be used. The above is Charles’ law: (V/T) is a constant when the pressure is constant.}

(b)

(i) Work done by gas = pΔV = (4.2×10⁵) ([3.87 – 3.49]×10³) ×10^-6 = 160J

(ii)

Increase / change in internal energy of gas = heating of system + work done on system

Increase / change in internal energy of gas = 565 – 160 = 405J

(c)

The internal energy is the sum of kinetic energy and potential energy / E_k + E_p. There are no intermolecular forces and so, no potential energy. (So, ΔU = ΔE_k)                          

Question 869: [Vectors]

Two forces of equal magnitude are represented by two coplanar vectors. One is directed eastwards and other is directed northwards.

What is the direction of a single force that will balance these two forces?

A towards the north-east

B towards the north-west

C towards the south-east

D towards the south-west

Reference: Past Exam Paper – November 2013 Paper 13 Q3

Solution 869:

Answer: D.

This question illustrates the need to write on the question paper.

Basically, this is only a simple vector addition of 2 vectors. The single force that would balance these 2 forces should be equal in magnitude and opposite in direction to the resultant of the 2 forces stated.

A vector diagram here would show how a balancing force must be pointing south-west.

Question 870: [Dynamics > Resultant force]

A hailstone, initially stationary at the base of a cloud, falls vertically towards Earth. Diagram shows the magnitudes and directions of the forces acting on the hailstone as it starts to drop.

Which diagram shows magnitudes and directions of these forces when the hailstone attains a terminal (constant) speed in the air (of uniform density)?

Reference: Past Exam Paper – November 2012 Paper 13 Q15

Solution 870:

Answer: D.

At terminal (constant) speed, the resultant force on the hailstone is zero such that the resultant acceleration of the hailstone is zero.

The gravitational weight of an object is always towards (downwards) the surface of the planet (Earth) and has a constant value near the Earth.

The ‘upthrust U’ should be recognised as being the result of the density of air. It is not just an upward force. The pressure at the bottom of the hailstone is slightly greater than that at the top. This causes a force on the surface of the hailstone. (Pressure = Force / Area)

Pressure P = hρg. The difference in height (h) between the top and the bottom of the hailstone is constant. Thus, the resultant pressure and the resultant force on it will also be constant. That is, the upthrust is constant.

The viscous force is the drag force of the air (similar to a frictional force) and this increases until a constant speed is reached with zero resultant force. This is answer D.