Thermodynamics Tricky Questions Part 2: Thermodynamics is a topic of Physics that deals with the interaction of one body with another in terms of quantities of heat and work. JEE Mains students must be aware of four laws of thermodynamics, i.e. Zeroth Law of Thermodynamics, First Law of Thermodynamics, Second Law of Thermodynamics, and Third Law of Thermodynamics.

JEE Mains examination consists of Physics, Chemistry and Mathematics, and to clear the examination; students must work very hard and with utmost dedication as it is a very competitive examination. In this article, we explain Thermodynamics, the laws of Thermodynamics and a few tricky questions that you need to be aware of to score good marks.

What is Thermodynamics?

Thermodynamics is a branch of physics that deals with heat, work and temperature, and their relation to energy, radiation and physical properties of matter. Let us take a look at the 4 Laws of Thermodynamics.

Zeroth Law of Thermodynamics

Zeroth Law of Thermodynamics states that when two bodies which are in thermal equilibrium with a third body are in thermal equilibrium with each other.

First Law of Thermodynamics

First Law of Thermodynamics states that, heat is a form of energy and it obeys the law of conservation of energy. This means that heat energy cannot be created or destroyed. It can, however, be transferred from one location to another and converted to and from other forms of energy.

Second Law of Thermodynamics

Second Law of Thermodynamics states that, any spontaneously occurring process will lead to an increase in the entropy of the universe. In other words, the entropy of an isolated system will never decrease over time. In some cases where the system is in thermodynamic equilibrium or going through a reversible process, the total entropy of a system and its surroundings remain constant. The second law is also known as the Law of Increased Entropy.

Third Law of Thermodynamics

Third Law of Thermodynamics states that, the entropy of a perfect crystal is zero when the temperature of the crystal is equal to absolute zero (0 K).

Thermodynamics Tricky Questions Part 2

Based on the previous year question papers, we have included a few thermodynamics based questions that are a tricky to answer and similar questions are expected to be asked in the examination.

Q.1: 200 g water is heated from 40° C to 60 °C. Ignoring the slight expansion of water, the change in its internal energy is close to (Given specific heat of water = 4184 J/kg/K)

(a) 167.4 kJ

(b) 8.4 kJ

(c) 4.2 kJ

(d) 16.7 kJ

Solution:

For isobaric process, ΔU = Q = msΔT

Here, m = 200 g = 0.2 Kg, s = 4184 J/Kg/K

ΔT = 60 ⁰C – 40 ⁰C = 20 ⁰C

ΔU = (0.2 )(4184)(20) =16736 J = 16.7 kJ

Answer: (d)16.7 kJ

Q.2: The work of 146 kJ is performed in order to compress one-kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7ºC. The gas is (R = 8.3 J mol^–1 K^–1)

(a) monoatomic

(b) diatomic

(c) triatomic

(d) a mixture of monoatomic and diatomic

Solution:

According to the first law of thermodynamics

ΔQ = ΔU + ΔW

For an adiabatic process, ΔQ = 0

0 = ΔU + ΔW

ΔU = -ΔW

nC_vΔT = -ΔW

C_v = -ΔW/nΔT = -[-146 x 10³]/[(1 x 10³) x 7] = 20.8 Jmol^-1K^-1

For diatomic gas, C_v = (5/2)R = (5/2)x8.3 = 20.8 Jmol^-1K^-1

Hence, the gas is diatomic

Answer: (b) diatomic

Q.3: From the following statements, concerning ideal gas at any given temperature T, select the correct.

(a)The coefficient of volume expansion at constant pressure is the same for all ideal gases

(b)The average translational kinetic energy per molecule of oxygen gas is 3kT, k being Boltzmann constant

(c)The mean-path of molecules increases with an increase in the pressure

(d)In a gaseous mixture, the average translational kinetic energy of the molecules of each component is different

Solution:

γ = dV/(V₀ x dT) at a constant temperature

γ = 1/V₀(dV/dT)_p since PV = RT

PdV = RdT or (dV/dT) = R/P₀

Therefore, γ = (1/V₀)(R/P₀) = R/RT₀

γ = 1/T₀

γ = 1/273

Answer: (a)The coefficient of volume expansion at constant pressure is the same for all ideal gases

Q.4: Calorie is defined as the amount of heat required to raise the temperature of 1 g of water by 1 °C and it is defined under which of the following conditions?

1. From 14.5 °C to 15.5 °C at 760 mm of Hg
2. From 98.5 °C to 99.5 °C at 760 mm of Hg
3. From 13.5 °C to 14.5 °C at 76 mm of Hg
4. From 3.5 °C to 4.5 °C at 76 mm of Hg

Solution:

1 calorie is the amount of heat required to raise the temperature of 1gm of water from 14.5 ⁰C to 15.5 ⁰C at 760 mm of Hg

Answer: (a) From 14.5 °C to 15.5 °C at 760 mm of Hg

Q.5: The average translational kinetic energy of 02 (molar mass 32) molecules at a particular temperature is 0.048 eV. The translational kinetic energy of N2 (molar mass 28) molecules in eV at the same temperature is

(a) 0.0015

(b) 0.003

(c) 0.048

(d) 0.768

Solution:

Average Kinetic Energy = (3/2)KT

It depends on temperature and does not depend on molar mass

For both the gases, average translational kinetic energy will be the same ie.,0.048 eV

Answer: (c) 0.048

Q.6: One mole of a monoatomic gas is heated at a constant pressure of 1 atmosphere from OK to 100K. If the gas constant R=8.32 J/mol K, the change in internal energy of the gas is approximately [1998]

(a) 2.3 J

(b) 46 J

(c) 8.67 × 10³J

(d) 1.25 x l0³J

Solution:

ΔU = nC_vdT =1x (3R/2)ΔT

ΔU = (3/2) x (8.3) x (100) = 1.25 x 10³ J

Answer: (d) 1.25 x l0³J

Q.7: An ideal gas heat engine is operating between 227 °C and 127 °C. It absorbs 10⁴J of heat at a higher temperature. The amount of heat converted into work is

(a) 2000 J

(b) 4000 J

(c) 8000 J

(d) 5600 J

Solution:

η =1- (T₂/T₁)

η =1- (127 + 273)/(227 + 273) = 1 – (400/500) = ⅕

W = ηQ₁ = ⅕ x 10⁴ = 2000 J

Answer: (a) 2000 J

So, these are a few questions that you should study and know how to solve them using the laws and formulas of thermodynamics.

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We hope this article on Thermodynamics Tricky Questions Part 2 has given you a few important questions that you need to be prepared before attending the JEE Mains examination. Stay tuned to embibe.com for more such informative articles.