Thermodynamics- AIPMT 2009 Multiple Choice Questions

Here are two multiple choice questions on thermodynamics which appeared in AIPMT 2009 question paper:

(i) In thermodynamic processes which of the following statements is not true?

(1) In an isochoric process pressure remains constant

(2) In an isothermal process the temperature remains constant

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

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

The only statement which is not true is statement (1). A process in which the pressure remains constant is called isobaric process (and not isochoric process).

[Note that an isochoric process is one in which the volume remains constant].

(ii) 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

When a quantity Q of heat is supplied to a system it is used to do an amount of work W by the system and to increase the internal energy of the system by ∆U:

Q = ∆U + W

Note that Q given in kilo calories is to be converted into joule which is the SI unit of heat. Therefore Q = 2×1000×4.2 J = 8400 J (remembering that 1 cal = 4.2 J very nearly).

The increase the internal energy, ∆U = Q – W =8400 – 500 = 7900 J.

Find some more multiple choice questions (with solution) in this section here as well as here.

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.

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.