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.

 

IIT-JEE Multiple Correct Answer Type Questions on Thermodynamics

"Hate the sin, love the sinner."
–Mahatma Gandhi

Today we will discuss two multiple correct answer type questions on thermodynamics. These were included in the IIT-JEE 2009 question paper:

(1) The figure shows the P–V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,

(A) the process during the path A → B is isothermal

(B) heat flows out of the gas during the path B → C → D

(C) work done during the path A → B → C is zero

(D) positive work is done by the gas in the cycle ABCDA.

Statement (A) is incorrect since a curve representing an isothermal in a PV diagram must be a hyperbola and not an arc of a circle.

We have PV = nRT with usual notations. The value of V is smaller at D compared to that at B where as the value of P is the same. The product PV = nRT is therefore smaller at D. This means that the temperature at D is less than that at B and hence statement (B) is correct.

During the path AB the gas expands, doing work. During the path BC the gas is compressed and work is done on the gas. But these works are unequal since the area under AB is greater than the area under BC. Therefore, statement (c) is incorrect.

The cycle of operations ABCDA is clockwise and the cyclic curve encloses an area. Therefore, positive work is done by the gas and statement (D) is correct.

So the correct options are (B) and (D).

(2) C_V and C_P denote the molar specific heat capacities of a gas at constant volume and constant pressure respectively. Then

(A) C_P – C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(B) C_P + C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(C) C_P/C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas

(D) C_P.C_V is larger for a diatomic ideal gas than for a mono-atomic ideal gas.

In the case of ideal gases C_P – C_V = R, the universal gas constant. Therefore statement (A) is wrong.

C_P and C_V are larger for a diatomic ideal gas compared to a mono-atonic ideal gas. Therefore, C_P + C_V is larger for a diatomic ideal gas. Statement (B) is correct.

[For diatomic ideal gas C_P = 7R/2 and C_V = 5R/2. For mono-atomic ideal gas C_P = 5R/2 and C_V = 3R/2].

Obviously statement (C) is wrong.

Statement (D) is correct since both C_P and C_V are larger for a diatomic ideal gas than for a mono-atomic ideal gas.

Therefore, options (B) and (D) are correct.

You will find a useful post on thermodynamics here.