JEE (Main) 2014 and JEE (Main) 2013 Multiple Choice Questions on Thermodynamics

"I object to violence because when it appears to do good, the good is only temporary; the evil it does is permanent."

– Mahatma Gandhi

Today we will discuss the multiple choice single answer type questions on thermodynamics, which appeared in JEE (Main) 2014 and JEE (Main) 2013 question papers.

(1) One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement:

(1) The change in internal energy in the whole cyclic process is 250R

(2) The change in internal energy in the process CA is 700R

(3) The change in internal energy in the process AB is – 350R

(4) The change in internal energy in the process BC is – 500R

The change in the internal energy ΔU  is given by

            ΔU =  n C_v ΔT where n is the number  of moles of gas, C_v is the molar specific heat of the gas at constant volume and ΔT is the change in the temperature of the gas.

The correct option is (4) since the change in internal energy in the process BC is 1×(5R/2) ×(–200) which is equal to – 500R.

[Note that the molar specific heat of diatomic ideal gas at constant volume is (5/2)R where R is the universal gas constant]

 

(2) The above P-V diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is:

(1) P₀V₀

(2) (13/2) P₀V₀

(3) (11/2) P₀V₀

(4) 4 P₀V₀

The upward vertical portion (on the left of the P-V diagram) and the rightward horizontal portion (on the top of the P-V diagram) represent the extraction of heat by the gas from the source. In the first case, heat is extracted (from the source) at constant volume V₀ (isochoric process) and in the second case, heat is extracted (from the source) at constant pressure 2P₀ (isobaric process).

Heat absorbed (Q, let us say) from the source during the isochoric process is given by

            Q = C_vn ΔT where C_v is the molar specific heat of the gas at constant volume, ΔT is the change in the temperature of the gas and n is the number of moles of gas in the engine.

Since PV = nRT where R is universal gas constant, we have

            ΔT = (ΔP)V/nR = (2P₀ – P₀)V₀/nR

Heat absorbed from the source during the isochoric process is therefore given by

            Q = C_vn (2P₀ – P₀)V₀/nR = (3/2)P₀V₀ since C_v = (3/2)R for an ideal monoatomic gas.

Heat absorbed (Q’, let us say) from the source during the isobaric process is given by

            Q’ = C_pn ΔT where C_p is the molar specific heat of the gas at constant pressure.

Since PV = nRT and the varying quantities are V and T, we obtain

            ΔT = P(ΔV)/nR = 2P₀(2V₀ – V₀)/nR

Heat absorbed from the source during the isobaric process is therefore given by

            Q’ = C_pn×2P₀(2V₀ – V₀)/nR

Since  C_p = (5/2)R for an ideal monoatomic gas, we have

            Q’ = 5P₀V₀

The total amount of heat extracted from the source in a single cycle is therefore given by

            Q₁ = Q + Q’ = (3/2)P₀V₀ + 5P₀V₀ = (13/2)P₀V₀

            You can access all questions on thermodynamics posted on this site by clicking on the label ‘thermodynamics’ below this post.

Essential points to be remembered for working out questions in thermodynamics can be seen here.

 

Heat engine and Refrigerator

From the section Thermodynamics, you will usually get questions on heat engines or refrigerators. You will definitely remember the expression for the efficiency (η) of a Carnot engine: η = (Q₁ –Q₂)/Q₁ = (T₁–T₂)/T₁ where Q₁ is the heat absorbed from the source at temperature T₁ and T₂ is the heat rejected to the sink at the lower temperature T₂.
The coefficient of performance (β) of a refrigerator (heat pump) is given by
β = (Heat removed from the cold body) / (Work done by the pump)
= Q₂/W = Q₂/(Q₁– Q₂) = T₂/(T₁–T₂)Now consider the following simple question which appeared in AIEEE 2002 question paper:
Even Carnot engine cannot give 100% efficiency because we cannot
(a) prevent radiation (b) find ideal source
(c) reach absolute zero temperature (d) eliminate frictionThe correct option is (c). The efficiency is given by the expression, η = (T₁–T₂)/T₁. The percentage efficiency is [(T₁–T₂)/T₁] ×100. This shows that the efficiency is 100% only if either the source temperature T₁ is infinite or the sink temperature T₂ is zero. Both are impossibilities.
Now see the following MCQ:
In a Carnot engine 800 J of heat is absorbed from a source at 400 K and 640 J of heat is rejected to the sink. The temperature of the sink is
(a) 320 K (b) 100 K (c) 273 K (d)250 K (e) 200 KIn a Carnot engine, Q₁/T₁ = Q₂/T₂ so that the temperature of the sink, T₂ = T₁Q₂/Q₁ = 400×640/800 = 320 K.
The following question appeared in Kerala Engineering Entrance 2000 question paper:
A heat engine undergoes a process in which its internal energy decreases by 400 J and it gives out 150 J of heat. During the process
(a) it does 250 J of work and its temperature rises
(b) it does 250 J of work and its temperature falls
(c) it does 550 J of work and its temperature rises
(d) it does 550 J of work and its temperature falls
(e) 250 J of work is done on the systemThe internal energy of the system will decrease when the system does work and/or gives off heat. Since the heat given out is 150 J and the reduction in internal energy is 400 J, the work done by the engine is 400– 150 = 250 J.
When the internal energy is reduced, the system is cooled. So, the correct option is (b).
Now, consider the following question:
The temperature inside a refrigerator is 4°C and the room temperature is 27°C. How many joules of heat will be delivered to the room for each joule of electricity consumed by the refrigerator?( Treat the refrigerator as ideal).
(a) 1 J (b) 12 J (c) 8.3 J (d) 13 J (e) 6 JThe coefficient of performance of the refrigerator, β = Q₂/W = Q₂/(Q₁– Q₂) = T₂/(T₁–T₂) = 277/(300–277) = 12. [Note that we have converted the temperature to the Kelvin scale].
Therefore, Q₂ =12 W. Heat delivered to the room is Q1 = Q₂+W = 12W+W = 13W. Here W is the work done by the pump. So for each joule of work done (for each joule of electricity consumed), the quantity of heat pumped out in to the room will be 13 joule.

Given below is a question of the type which often finds a place in Entrance test papers:

An ideal gas is taken through a cycle of operations shown by the indicator diagram. The net work done by the gas at the end of the cycle is

(a) 6P0V0 (b) 4P0V0 (c) 15P0V0 (d) 10P0V0 (e) 3P0V0

The work done in a cyclic process indicated by a PV diagram is the area enclosed by the closed curve. The area under the slanting curve showing the expansion of the gas from volume 2V0 to volume 5V0 gives the work done by the gas. This is greater than the area under the (horizontal) curve showing the compression of the gas (from volume 5V0 to volume 2V0), which gives the work done on the gas. The vertical portion of the curve is an isochoric (volume constant) change which involves no work since the area under it is zero. The area enclosed by the closed curve gives the net work done by the gas. The triangular area enclosed is ½ ×3V0×2P0 = 3P0V0 [Option (e)].
[ Note that in problems of the above type, the work is done by the gas if the arrow showing the cycle is clockwise. If the arrow is anticlockwise, work is done on the gas. In either case, the work done is the area enclosed by the curve].