Multiple Choice Practice Questions on AP Physics B Thermodynamics

“This time, like all times, is a very good one if we but know what to do with it”

– Ralph Waldo Emerson

You will find many practice questions (with solution) on thermodynamics on this site. You can access them by clicking on the label ‘thermodynamics’ below this post. Today we shall discuss a few more multiple choice practice questions in this section for the benefit of AP Physics B aspirants.

(1) Keeping the pressure constant, the temperature of m kg of a gas is raised through ΔT º C. If the specific heats of the gas at constant volume and constant pressure are C_v and C_p respectively, the increase in internal energy of the gas is

(a) m C_p ΔT

(b) m (C_p – C_v) ΔT

(c) m C_v ΔT

(d) m (C_p + C_v) ΔT

(e) m (C_p + C_v) ΔT/2

If the temperature increase occurs at constant volume, the entire energy supplied to the gas is used up in increasing the internal energy of the gas. The energy required for increasing the temperature of one kilogram of the gas through 1º C at constant volume is the specific heat of the gas at constant volume (C_v). When the temperature increase occurs at constant pressure, the gas has to expand and therefore has to do external work. The specific heat of the gas at constant pressure is equal to the energy supplied for increasing the temperature of one kilogram of the gas through 1º C at constant pressure. But the increase in the internal energy of 1 kg of the gas on heating through 1º C in this case also is C_v.

Therefore, the increase in internal energy (ΔU) of m kg of the gas when its temperature is raised through ΔT º C is given by

             ΔU = m C_v ΔT

(2) In the above question the work done by the gas is

(a) mC_pΔT

(b) m(C_p – C_v)ΔT

(c) mC_vΔT

(d) m(C_p + C_v)ΔT

(e) Zero

Keeping the pressure constant, when the temperature of m kg of the gas is raised through ΔT º C, the total energy required is mC_pΔT. Out of this the energy used in increasing the internal energy of the gas is mC_vΔT. The energy spent for doing work is therefore given by the difference between the two values and is equal to m(C_p – C_v)ΔT.

(3) A sample of gas contained in a cylinder undergoes a cyclic process shown by the adjoining PV diagram. Among the following statements which one is correct?

(a) Work is done on the gas during the process shown by AB

(b) Work done by the gas during the cycle ABCA is negative

(c) BC represents an isochoric  process

(d) CA represents an isobaric process

(e) Work done on the gas during the process CA is zero

During the process shown by AB the gas expands. Therefore work is done by the gas (and not on the gas). Statement (a) is therefore incorrect.

Since the cycle ABCA is clockwise, work done by the gas is positive. Statement (b) is therefore incorrect.

[Work done by the gas during its expansion represented by the area under the curve AB is greater than the work done on the gas during its contraction represented by the area under the curve BC]

During the process BC the volume of the gas changes and hence the process is not isochoric.

[Isochoric process is also known as isovolumetric process]

During the process CA the pressure of the gas changes and hence the process is not isobaric.

Work done on the gas uring the process CA is zero since CA represents an isochoric process. Therefore option (e) is correct.

[Note that work is done only if volume changes].

(4) A Carnot engine operates using a high temperature source at 400 K and a low temperature sink at 300 K. How much more efficient will this engine be if the temperature of the sink is reduced to 200 K?

(a) Twice as efficient

(b) Three times as efficient

(c) Four times as efficient

(d) Five times as efficient

(e) Six times as efficient

Efficiency (η) of Carnot engine is given by

            η = (Q₁ – Q₂)/Q₁ = (T₁ – T₂)/T₁ where Q₁ is the quantity of heat absorbed from the source at the source temperature T₁ and Q₂ is the quantity of heat liberated to the sink at the sink temperature T₂.

The efficiency η₁ when the source and sink are at 400 K and 300 K respectively is given by

            η₁ = (400 – 300)/400 = ¼

The efficiency η₂ when the source and sink are at 400 K and 200 K respectively is given by

             η₂ = (400 – 200)/400 = ½

The engine is therefore twice as efficient [Option (a)].

(5) A fixed mass of gas does 25 J of work on its surroundings and transfers 15 J of heat to the surroundings. The internal energy of the gas

(a) decreases by 15 J

(b) increases by 15 J

(c) remains unchanged

(d) decreases by 40 J

(e) increases by 40 J

The internal energy of the gas decreases when it does work on the surroundings and also when it transfers heat to the surroundings. The total decrease in the internal energy of the gas is therefore equal to 25 J + 15 J = 40 J [Option (d)].

You will find similar questions with solution here.

AP Physics B – Answer to Free Response Practice Question on Thermodynamics

“Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time”

– Thomas A. Edison

A free response practice question on thermodynamics was posted on 24^th October 2010 for AP Physics B aspirants. As promised, I give below a model answer (along with the question) for your benefit.

The adjoining figure shows a fixed cylindrical vessel of inner diameter 10 cm fitted with a smooth, light piston connected to a light spring of spring constant 4000 Nm^–1. The other end of the spring is attached to an immovable support. The cylinder contains an ideal gas at 27º C. The spring is initially in its released condition and the initial volume of the gas in the cylinder is 0.8×10^–3 m³. When heat is supplied to the gas, it expands and pushes the piston through 10 cm. The atmospheric pressure is 10⁵ pascal. Now answer the following questions:

(a) What are the initial pressure and the initial absolute (Kelvin) temperature of the gas?

(b) Calculate the potential energy acquired by the spring because of the expansion of the gas.

(c) Calculate the final pressure of the gas in the cylinder.

(d) Calculate the final temperature of the gas in the cylinder.

(a) The initial pressure P₁ of the gas is the same as the atmospheric pressure 10⁵ Nm^–1. The initial temperature T₁ of the gas is (27 + 273) K = 300 K.

(b) The potential energy of the spring is U = ½ kx² where k is the spring constant and x is the compression of the spring.

Therefore, U = ½ ×4000×(0.1)² = 20 J.

(c) The final pressure of the gas is the sum of the atmospheric pressure and the pressure due to the elastic force developed in the spring.

Elastic force = kx.

Since this force acts over the area A of the piston, pressure due to the elastic force is kx/A = kx/πR² where R is the radius of the piston.

Therefore, final pressure P₂ of the gas = 10⁵ + (4000×0.1)/(π×0.05²) = 10⁵ + 0.51×10⁵ = 1.51×10⁵ pascal.

(d) The final temperature T₂ of the gas is given by

P₁V₁/T₁ = P₂V₂/T₂

Therefore, T₂ = (P₂V₂ T₁)/ (P₁V₁)

The final volume V₂ of gas = Initial volume + πR²x = 0.8×10^–3 + (π×0.05²×0.1) = 0.8×10^–3 + 0.79×10^–3 = 1.59×10^–3 m³

Therefore, final temperature T₂ = (1.51×10⁵×1.59×10^–3 ×300)/(10⁵×0.8×10^–3) = 900 K.

AP Physics B – Free Response Practice Question on Thermodynamics

Today I give you a free response practice question (for AP Physics B) on thermodynamics. You can access all posts on thermodynamics on this site by clicking on the label ‘thermodynamics’ below this post or by trying a search using the search box provided on this page.

Here is the question:

The adjoining figure shows a fixed cylindrical vessel of inner diameter 10 cm fitted with a smooth, light piston connected to a light spring of spring constant 4000 Nm^–1. The other end of the spring is attached to an immovable support. The cylinder contains an ideal gas at 27º C. The spring is initially in its released condition and the initial volume of the gas in the cylinder is 0.8×10^–3 m³. When heat is supplied to the gas, it expands and pushes the piston through 10 cm. The atmospheric pressure is 10⁵ pascal. Now answer the following questions:

(a) What are the initial pressure and the initial absolute (Kelvin) temperature of the gas?

(b) Calculate the potential energy acquired by the spring because of the expansion of the gas.

(c) Calculate the final pressure of the gas in the cylinder.

(d) Calculate the final temperature of the gas in the cylinder.

The above question carries 10 points. You have about 11 minutes for answering it.

Try to answer the question. I’ll be back soon with a model answer for your benefit.

AP Physics B - Multiple Choice Practice Questions on Kinetic Theory of Gases

Essential points to be remembered in kinetic theory of gases were discussed in the post dated 13^th March 2008. Questions on kinetic theory of gases were discussed subsequently. You can access all posts related to kinetic theory of gases by clicking on the label, ‘kinetic theory’ below this post. To access older posts you need to click on the ‘older posts’ button.

Today we will discuss a few more typical multiple choice questions on kinetic theory of gases:

(1) The root mean square (R.M.S.) speed v of the molecules of an ideal gas is given by the expressions,

v = √(3RT/M ) and

v = √(3kT/m ) where R is universal gas constant, T is the absolute (Kelvin) temperature, M is the molar mass, k is Boltzman’s constant and m is the molecular mass. The R.M.S. speed of oxygen molecules (O₂) at temperature T₁ is v₁. When the temperature is doubled, if the oxygen molecules are dissociated into atomic oxygen, what will be R.M.S. speed of oxygen atoms? (Treat the gas as ideal).

(a) v₁/2

(b) v₁

(c) √2 v₁

(d) 2v₁

(e) 4v₁

We have v₁ = √(3RT₁/M ) or

v₁ = √(3kT₁/m )

On dissociation the molar mass as well as the molecular mass gets halved. Using the second equation, the R.M.S. speed v after dissociation is given by

v = √[3k×2T₁/ (m/2 )] = 2√(3kT₁/m ) = 2v₁

(2) Four moles of an ideal diatomic gas is heated at constant volume from 20º C to 30º C. The molar specific heat of the gas at constant pressure (C_p) is 30.3 Jmol^–1K^–1 and the universal gas constant (R) is 8.3 Jmol^–1K^–1. The increase in internal energy of the gas is

(a) 80.3 J

(b) 303 J

(c) 332 J

(d) 880 J

(e) 1212 J

The increase in internal energy is MC_v∆T where M is the mass of the sample of the gas, C_v is the specific heat at constant volume and ∆T is the rise in temperature of the gas. If we use the molar specific heat of the gas at constant volume for C_v, the number of moles in the sample of the gas is to be used in the place of M.

Now, C_v = C_p – R = 30.3 – 8.3 = 22 Jmol^–1K^–1.

Therefore, the increase in internal energy of the gas is 4×22×10 = 880 J.

(3) In the case of real gases, the equation of state, PV = RT (where P, V and T are respectively the pressure, volume and absolute temperature), is strictly satisfied only if corrections are applied to the measured pressure P and the measured volume V. The corrections for P and V arise respectively due to

(a) intermolecular attraction and the size of molecules

(b) size of molecules and expansion of the container

(c) expansion of the container and intermolecular attraction

(d) kinetic energy of molecules and collision of molecules

(e) intermolecular attraction and collision of molecules

In kinetic theory of gases it is assumed that there is no force between molecules But there is actually intermolecular attraction which reduces the pressure. So the correction for P arises due to intermolecular attraction.

The entire volume V of the container is not available for the molecules since the molecules have a finite size. The assumption (in kinetic theory) that the molecules are point masses without appreciable volume is incorrect. So the correction for V arises due to the size of molecules.

The correct option is (a).

(4) Gases exert pressure on the walls of the container because the gas molecules

(a) collide one another

(b) exert intermolecular attraction

(c) possess momentum

(d) expand on absorbing heat

(e) exert repulsive force

Because of the momentum of the gas molecules, they collide with the walls of the containing vessel and momentum transfer takes place, resulting in a force on the walls. Pressure is force per unit area. The basic reason for the pressure is the momentum of the gas molecules [Option (c)].

Now, see similar questions with solution here.

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 –mv – mv = – 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 = P’V₁ and 4P×V = P’V₂

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 = (Q₁ – Q₂)/Q₁ = (T₁– T₂)/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 Q₁ – Q₂ 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 = mc∆T

Therefore, ∆T = gh/c

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