Thermodynamics – Two Multiple Choice Questions

Often simple questions on thermodynamics may cause unexpected confusion. The reason for the confusion is usually the lack of your understanding of the fundamentals. If you don’t have any confusion in respect of the following questions, well and good!

(1) A sample of an ideal gas initially having internal energy U₁ and pressure P₁ expands adiabatically and performs work W. Heat energy Q is then added to the gas at constant volume so that its pressure is increased to the initial value P₁. As a result of the above processes, the internal energy of the gas

(a) decreases by Q – W

(b) increases by Q – W

(c) decreases by Q

(d) increases by Q

(e) remains unchanged

Since the gas expands adiabatically and outputs mechanical energy, the internal energy of the gas is decreased. [Remember that during adiabatic process there is no heat transfer between the gas and the surroundings]. On adding heat energy to the gas at constant volume the internal energy of the gas is increased. There is no work involved since the volume is constant (isochoric process).

Evidently the internal energy of the gas increases by Q – W.

[You will obtain the answer from the mathematical statement of the 1^st law of thermodynamics: ∆Q = ∆U + ∆W where ∆Q is the heat energy supplied to the system by the surroundings, ∆W is the work done by the system on the surroundings and ∆U is the increase in the internal energy of the system].

(2) In the given PV diagram, I is the initial state and F is the final state. The gas goes from I to F by (i) IAF (ii) IBF (iii) ICF. The heat absorbed by the gas is

(a) the same in all three processes

(b) the same in (i) and (ii)

(c) greater in (i) than in (ii)

(d) the same in (i) and (iii)

(e) greater in (iii) than in (ii)

This question appeared in Kerala engineering entrance 2009 question paper.

We have ∆Q = ∆U + ∆W

Since the same initial point (I) and the same final point (F) are given for all the three paths, the change in the internal energy (∆U) is the same for all the three processes. During the path IAF the net work done by the gas (∆W) is positive since the work done on the gas during the compression IA (given by the area under the line IA) is less than the work done by the gas during its expansion AF. No work is involved during the process IBF since the volume is constant (isochoric process). During the path ICF the net work done (∆W) by the gas is negative since the work done by the gas during its expansion IC is less than the work done on the gas during its compression CF.

Therefore, the heat absorbed by the gas (∆Q) is greater in (i) than in (ii).

You will find useful multiple choice questions with solution at apphysicsresources as well as at physicsplus.

All India Pre-Medical/Pre-Dental Entrance Examination (Preliminary) 2009 (AIPMT 2009) Questions from Thermodynamics

The following questions were included from thermodynamics in the All India Pre-Medical/Pre-Dental 2009 Entrance Examination (Preliminary):

1. The internal energy change in a system that has absorbed 2 Kcals of heat and done 500 J of work is

(1) 6400 J

(2) 5400 J

(3) 7900 J

(4) 8900 J

We have ∆Q = ∆U + ∆W where ∆Q is the heat supplied to the system, ∆U is the increase in internal energy of the system and ∆W is the work done by the system.

Therefore, the increase in internal energy of the system is given by

∆U = ∆Q – ∆W = 2000×4.2 J – 500 J = (8400 – 500) J = 7900 J

[Note that 1 calorie = 4.2 joule, nearly]

2. In thermodynamic process which of the following statements is not true?

(1) In an isochoric process pressure remains constant.

(2) In an isothermal process temperature remains constant.

(3) In an adiabatic process PV^γ = constant.

(4) In an adiabatic process the system is insulated from the surroundings.

Statement (1) is not true since an isochoric process is one in which volume remains constant. So option (1) is the answer.

You will find many useful questions on thermodynamics at physicsplus here and at other locations.

You may click on the label ‘thermodynamics’ below relevant posts to access all posts related to thermodynamics. Or, you may try a search using the ‘search blog’ box at the top of the site.

Two AIPMT 2008 Questions (MCQ) from Current Electricity

Some of you may understand the principles in physics very well but your capacity for numerical manipulations may be poor. Practice can make you strong in solving questions involving numerical manipulations so that you will not waste your precious time on such questions. Here are two questions which appeared in AIPMT 2008 question paper:

(1) An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20º C? The temperature of boiling water is 100º C.

(1) 8.4 min

(2) 12.6 min

(3) 4.2 min

(4) 6.3 min

We have VIt =mcθ where V is the voltage, I is the current, t is the time of flow of the current, m c is the specific heat of water (which is approximately 4200 Jkg^–1 K^–1) and θ is the temperature rise. is the mass of water,

Therefore, 220×4×t = 1×4200×(100 – 20)

This will give t = 381 sec. (nearly) which is approximately 6.3 min.

(2) a galvanometer of resistance 50 Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be

(1) 5550 Ω

(2) 6050 Ω

(3) 4450 Ω

(4) 5050 Ω

Since the current through an ordinary galvanometer is directly proportional to the deflection (remember that in a tangent galvanometer this is not the case) we have

3/(50+2950) = k×30 when the deflection is 30 divisions.

Here k is the proportionality constant (figure of merit of the galvanometer).

If the resistance in series for reducing the deflection to 20 divisions is X we have

3/(50+X) = k×20

On dividing the first equation by the second,

(50+X)/3000 = 3/2 from which X = 4450 Ω.

Two AIPMT 2008 Questions from Thermal Physics

The following questions appeared in the All India Pre-Medical / Pre-Dental Entrance 2008 Examination:

(1) At 10º C the value of the density of a fixed mass of an ideal gas divided by its pressure is x. At 110º C this ratio is

(1) 383x/283

(2) 10x/110

(3) 283x/383

(4) x

We have PV = rT where r is the gas constant for the given mass.

Since the volume V = M/ρ where M is the mass and ρ is the density, we have

P/ρ = rT/M

Since r and M are constants for a given mass of gas, we have

(ρ₁/P₁) /(ρ₂/P₂) = T₂/T₁ where the suffix 1 is for the quantities at temperature 10º C and suffix 2 is for the quantities at temperature 110º.

It is given that (ρ₁/P₁) = x.

Therefore x/(ρ₂/P₂) = 383/283 since the temperatures are in degree kelvin.

From this (ρ₂/P₂) = 283x/383

(2) On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39º W and 239º W respectively. What will be the temperature on the new scale, corresponding to a temperature of 239º C on the Celsius scale?

(1) 117º W

(2) 200º W

(3) 139º W

(4) 78º W

We have 0º C = 39º W and 100º C = 239º W

A temperature difference of 100º C is therefore equivalent to a temperature difference of 200º W. A temperature difference of 1º C is thus equivalent to a temperature difference of 2º W.

Therefore, a temperature 39º C is equal to [39 + (39×2)]º W = 117º W

MCQ on Wien’s Law

The following question on Wien’s law is a typical one and is of the type often found in Medical and Engineering Entrance test papers:

A black body emits radiations of maximum intensity for the wave length of 7000Ǻ, when the temperature of the body is 3927ºC. If the temperature of the body is increased by 1000ºC, the maximum intensity would be observed at nearly

(a) 3650Ǻ (b) 4650Ǻ (c) 5650Ǻ

(d) 6000Ǻ (e) 6650Ǻ

The wave length λ_m corresponding to the maximum intensity emission from a black body at a temperature T is given by Wien’s law,

λ_mT = constant [= 0.29 cm K, if you express λ_m in cm].

In other words, λ_m1T₁ = λ_m2T₂

[Remember, you have to substitute the absolute (Kelvin) temperatures in this equation].

Therefore, 7000×4200 = λ_m2×5200, from which λ_m2 =5650Ǻ, nearly.

[For question setters: When a question is set using Wien’s law, it should be remembered that the product λ_mT satisfies the condition, λ_mT = 0.29 cmK, if λ_m is expressed in cm. I write this because I have seen questions in which this point is ignored]

Questions from Thermodynamics

(1) In the cyclic process shown in the PV diagram for an ideal gas, the net work done by the gas during one cycle is
(a) 24P₀ V₀ (b) –24P₀V₀ (c) – 6πP₀V₀ (d) 6πP₀V₀ (e) 4πP₀V₀

The area enclosed by the closed curve gives the work done. Since the cycle of operations is clockwise (in the PV diagram with P on the Y-axis), the net work done by the gas is positive

Therefore net work done by the system during one cycle = Area of circle.
Since the quantities on the two axes are different, the shape is not rally a circle. The scales of the quantities could have been changed to make it an ellipse. So, treat it as an ellipse and calculate the area as πab where ‘a’ and ‘b’ are the semi-major and semi-minor axes respectively.
So, work done = π[(5P₀ –2P₀)/2] [(10V₀ –2V₀)/2] = 6πP₀V₀
[Note that if the arrow in the diagram shows the processes in the anticlockwise direction, the net work done by the system will be negative].
(2) A refrigerator removes 1000 calories of heat from the ice trays. The coefficient of performance of the refrigerator is 10. Then the work done by the compressor motor is at least
(a) 240 J (b) 420 J (c) 4200 J (d) 42000 J (e) 0.042 J
Coefficient of performance, β = Q2/w where Q₂ is the quantity of heat removed from the cold body (inside the refrigerator) and ‘w’ is the work needed to transfer this heat to the hot body (outside).
Therefore w = Q/β = 1000/10 = 100 calories = 420 J, approximately. This is the minimum work to be done by the compressor motor.
(3) An engine takes in 10000 J of heat and rejects 8000 J while operating between temperatures of 900 K and 600 K. The actual efficiency is
(a) 1/5 (b) 2/5 (c) 1/3 (d) 2/3 (e)4/5
This simple question appeared in the Kerala Engineering Entrance 2000 test paper. The actual efficiency is given by η = (Q₁ –Q₂)/Q₁. Note that this will be less than (T₁–T₂)/T₁ in practical heat engines. The efficiency given by the two expressions will be equal in the case of the ideal, perfectly reversible Carnot engine only.
Actual efficiency = (10000–8000)/10000 = 1/5.

Kerala Medical Entrance Test 2006- Two Questions on Heat

The following two questions appeared in the Kerala Medical Entrance 2006 test paper:
(1) A thermos flask made of stainless steel contains several tiny lead shots. If the flask is quickly shaken up and down several times, the temperature of lead shots
(a) increases by adiabatic process (b) increases by isothermal process (c) decreases by adiabatic process (d) remains the same (e) first decreases and then increases.
This is a simple question aimed at testing your understanding of a basic concept. In an adiabatic process, there is no heat transfer to or from the system. The system in the question is the lead shots contained in the thermos flask which is thermally insulated from outside. The temperature of the lead shots is therefore increased by adiabatic process [Option (a)].
(2) A body cools from 50˚C to 49.9˚C in 5 s. How long will it take to cool from 40˚C to 39.9˚C? Assume the temperature of the surroundings to be 30˚C and Newton’s law of cooling to be valid.
(a) 2.5 s (b) 5 s (c) 20 s (d) 10 s (e) 15 s
As per Newton’s law of cooling, the rate of cooling of a body is directly proportional to the mean excess of temperature of the body over the surroundings (if the absolute temperature of the body is nearly equal to the absolute temperature of the surroundings).
Therfore, we have (50 – 49.9)/5 α (49.95 – 30) and
(40 – 39.9)/t α (39.95 – 30)
Dividing the first one by the second, we obtain t/5 = 2, very nearly, so that the required time for cooling is 10 s.