COMEDK UGET 2025 Thermodynamics Practice Questions with Solutions

COMEDK UGET 2025 Thermodynamics Practice Questions with Solutions: Thermodynamics is a chapter that is involved in both Physics as well as Chemistry subjects in the COMEDK UGET exam. To excel in both the subjects, you must boost your exam preparation by solving practice questions for COMEDK Thermodynamics. The COMEDK UGET 2025 Thermodynamics practice questions with solutions highlights some of the most probable questions and important topics frequently asked or previously appeared in the exam paper. By attempting these sample questions you will have a better chance of scoring full marks in the entrance test. The Thermodynamics chapter also holds a significant weightage in both of the subjects. Where in Physics, the weightage ranges from 16-25%, in Chemistry, it has an expected weightage ranging from 8-25%. Thermodynamics also covers various subtopics such as the Laws of Thermodynamics, Heat Engine and Refrigerator, Thermodynamic Quantities etc. Around 5-15 questions in Chemistry and 10-15 questions in Physics paper are expected to appear from Thermodynamics in the COMEDK UGET 2025 exam.

The COMEDK syllabus 2025 covers topics from 11th and 12th standards, thereafter, it is advised that you practice COMEDK UGET Thermodynamics important questions consistently rather than studying all the topics from scratch. After you do this, you will be able to get a hold of your time management skills, question type, exam pattern etc. To get more details about the COMEDK UGET 2025 Thermodynamics practice questions with solutions, read this article.

Also Check - Do or Die Chapters for COMEDK UGET 2025 Chemistry

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COMEDK UGET 2025 Thermodynamics Important Topics

The chapter on Thermodynamics in COMEDK UGET holds a good volume of subtopics such as the First Law of Thermodynamics, Thermodynamic Terms, Thermochemistry etc. Being an exam taker, you must know that the questions in COMEDK UGET exam will be based on these topics. Therefore, you must pay heed to all these topics. The table below includes the gist of the list of COMEDK UGET 2025 Thermodynamics sub-topics for exam preparation:-

COMEDK UGET 2025 Thermodynamics Expected Weightage

If you're appearing for COMEDK UGET, it's essential to be aware of the weightage and the total number of questions related to the topic of Thermodynamics. This information will help you plan your preparation efficiently. The table below presents a concise overview of the COMEDK UGET Thermodynamics expected weightage and the estimated number of topics:-

Also Check - COMEDK Chapter Wise PYQs for Chemistry

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COMEDK UGET 2025 Thermodynamics Practice Questions with Solutions

Regularly solving COMEDK UGET Thermodynamics sample questions can greatly improve your exam preparation. Practicing with COMEDK Thermodynamics questions and solutions allows you to familiarize yourself with possible exam questions, verify correct answers, and grasp step-by-step solutions. Below are some key COMEDK UGET 2025 Thermodynamics practice questions, carefully curated by experts from past exam papers:

Q1. Which of the following are not path functions?

I) H-TS

II) W

III) q

IV) q+W

a) I & III

b) I, II, IV

c) II, III, IV

d) I & IV

Correct answer - d) I & IV

Explanation: Thermodynamic parameters which depend only on the initial and final states of system are called state functions, such as enthalpy (H = q + W), Gibb's free energy (G = H - TS).

Whereas, thermodynamic parameters which depend on the path by which the process is performed and not on initial and final states, are called path functions, such as work done (W), heat (q) etc.

Q2. In which of the following changes, entropy decreases?

a) Rusting of iron

b) Melting of ice

c) Vaporisation of camphor

d) Crystallisation of sucrose from solution 

Correct answer - d) Crystallisation of sucrose from solution

Explanation: Among the given options, entropy decreases during the crystallisation of sucrose from solution. It is because entropy is a measure of randomness, and during the process of crystallisation liquid state changes into solid state. Hence, entropy decreases.

Q3. 5.8g of a gas maintained at 95℃ occupies the same volume as 0.368g of hydrogen gas maintained at a temperature of 17℃ and pressure being the same atmospheric pressure for both the gases. What is the molecular mass of the unknown gas?

a) 44 g/ mol

b) 32 g/ mol

c) 71 g/mol

d) 40g/ mol

Correct answer - d) 40g/ mol

Explanation: To find the molecular mass of the unknown gas, we will use the Ideal Gas Law and the concept of molar mass. The Ideal Gas Law is given by:

PV = nRT

Where:

P is the pressure of the gas

V is the volume of the gas

N is the number of moles of the gas

R is the universal gas constant

T is the temperature of the gas in Kelvin

Let's start by converting the temperatures given in degrees Celsius to Kelvin:

T_unknown = 95℃ = 95 + 273.15 = 368.15 K

T_hydrogen = 17℃ = 17 + 273.15 = 290.15 K

According to the problem, the volume and the pressure of both gases are the same. Thus, from the Ideal Gas Law:

n_unknown RT_unknown = h_ydrogen RT_hydrogen

Since R is constant, it cancels out. Rearranging the equation, we get:

n_unknown T_unknown = n_ydrogen T_hydrogen

To find the moles of each gas, we use the relation between the mass and the molar mass:

N = m/Molar Mass

For hydrogen (H2), the molar mass is approximately 2g/mol.

n_unknown = 0.368 g/2g per mol = 0.184 mol

Substituting the values into the equation, we get:

m_unknown /M_unknown x 368.15K = 0.184 mol x 290.15K

Where m_unknown = 5.8 g and M_ydrogen is the molar mass of the unknown gas which we need to find.

Solving, we have:

5.8/M_unknown x 368.15 = 0.184 x 290.15

5.8 x 368.15/M_unknown = 53.3876

M_unknown = 5.8 x 368.15/53.3876 = 40 g/mol

Thus, the molecular mass of the unknown gas is approximately 40 g/mol, which corresponds to option D.

Q4. The change in the energy of system if 500 cal of heat energy are added to a system and system does 350 cal of work on the surroundings will be:

a) - 150 cal

b) + 150 cal

c) - 850 cal

d) + 850 cal

Correct answer - b) + 150 cal

Explanation: Heat absorbed,

q = 500 cal

Work done by the system, W = 350 cal

According to the first law of Thermodynamics,

????E = q+W = 500 + (-350) = +150 cal

Q5. Which of the following is an intensive property?

a) Temperature

b) Surface tension

c) Viscosity

d) All of the above

Correct answer - d) All of the above

Explanation: Temperature, surface tension and viscosity are intensive properties. They are independent of the quantity of matter.

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