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## Sunday, April 25, 2010

### AP Physics B- Additional Multiple Choice Practice Questions on Kinetic Theory and Thermodynamics

Often you will get simple questions to answer. Don’t be too impulsive while answering seemingly simple questions The intention of the question setter in many cases will be to check whether you have mastered the fundamental principles. She may make a question a little bit indirect and you may tend to commit mistakes. It will be a good practice to find time to check your answers, especially in the case of multiple choice questions. Here are some simple questions on kinetic theory and thermodynamics.

(1) In the gas equation, 3PV = RT where R is universal gas constant, V stands for

(a) volume of 3 kg of gas

(b) volume of 3 g of gas

(c) volume of 3 mole of gas

(d) volume of 1/3 kg of gas

(e) volume of 1/3 mole of gas

The gas equation for n moles is

PV = nRT

The gas equation given in the question is

PV = (1/3) RT

The equation is therefore for 1/3 mole and V stands for the volume of 1/3 mole of gas [Option (e)].

(2) Identical balls each of mass m and moving with the same velocity v hit a fixed surface normally and elastically. If there are n hits per second, what is the average force exerted (by the balls) on the surface?

(a) mnv/2

(b) mnv

(c) 2mnv

(d) mnv2/2

(e) 2mnv2

Since the collision is elastic, the balls will rebound with the same speed. The initial momentum of each ball is mv and the momentum after collision with the surface is mv, the negative sign indicating the reversal of direction. The change of momentum due to the collision is –mvmv = – 2mv. Each collision therefore transfers a momentum 2mv to the surface. Since there are n collisions per second, the momentum transferred per second to the surface (which is the rate of change of momentum) is 2mnv.

Rate of change of momentum is force according to Newton’s second law and hence the average force exerted (by the balls) on the surface is 2mnv [Option (c)].

(3) The root mean square speed of molecules of a gas in a container maintained at temperature T is 600 ms–1. If 25% of the gas molecules leak out at this temperature, the root mean square speed of the remaining molecules is

(a) 300 ms–1

(b) 600 ms–1

(c) 600√2 ms–1

(d) 1200 ms–1

(e) 2400 ms–1

The root mean square speed of gas molecules is directly proportional to the square root of absolute temperature. [This follows from the postulate of kinetic theory which states that the average kinetic energy of gas molecules is directly proportional to the absolute temperature].

Since the temperature remains constant, the root mean square speed is unchanged [Option (b)].

(4) A cylinder containing a gas is divided into two parts A and B by a tight fitting fixed piston (Fig.). The pressure and volume of gas in the two parts A and B respectively are (P, 2V) and (4P, V). If the piston is released so that it can slide freely under isothermal condition, then the volume of the gas in the two parts A and B are respectively

(a) V, 2V

(b) 1.5V, 1.5V

(c) V, 2V

(d) 2V, V

(e) 0.5V, 2.5V

When the piston is left free, it moves towards the left and the pressures in the two parts get equalized. If this common pressure is P’, we have (since the temperature is unchanged)

P×2V + 4P×V = P×3V, from which P’ = 2P

If V1 and V2 are the final volumes of gas in the parts A and B respectively, we have

P×2V = PV1 and 4P×V = PV2

Substituting for P’ we obtain V1 = V and V2 = 2V [Option (a)].

[You can solve the above problem in no time if you note that the product of pressure and volume of gas in part B is twice the product of pressure and volume in part A. Since the final pressure is the same, the volume of gas in part B has to be twice the volume in part A. Let us consider another example:

What would be the final volumes in parts A and B if the initial pressure- volume states were (2P, 3V) and (12P, 2V) respectively?

The pressure volume products are 6PV and 24PV. So the total volume of 5V is to be divided in the ratio 6:24, that is 1:4. We obtain V1 = 5V×1/5 = V and V2 = 5V×4/5 = 4V].

(5) A Carnot engine working between 300 K and 800 K has a work output of 1000 joule per cycle. The amount of heat absorbed by the engine from the source is

(a) 400 J

(b) 500 J

(c) 800 J

(d) 1200 J

(e) 1600 J

In the case of a Carnot engine we have the expression for efficiency h given by

h = (Q1Q2)/Q1 = (T1T2)/T1 where Q1 is the quantity of heat absorbed from the source at higher temperature T1 and Q2 is the quantity of heat liberated to the sink at lower temperature T2.

Since Q1Q2 is the work output, we have

1000/Q1 = 500/800 from which Q1 = 1600 J

(6) The height of a water fall is h metre. Assume that the entire energy due to the fall is converted into heat. If the specific heat capacity of water is c Jkg–1K–1 and the acceleration due to gravity is g ms–2, the temperature difference between the top and bottom of the fall in degree Kelvin is

(a) c/gh

(b) gh/c + 273

(c) gh/c

(d) cgh

(e) cgh + 273

Water loses gravitational potential energy while falling down and gains an equivalent thermal energy, resulting in a rise of temperature. Considering a mass m of water, if T is the rise in temperature, we have

mgh = mcT

Therefore, ∆T = gh/c

You will find some useful multiple choice questions (with solution) in this section here.