Practice questions on KTG and Thermodynamics for JEE Advanced

15 Practice questions on KTG and Thermodynamics  for JEE Advanced

Are you ready to challenge yourself with some of the toughest Kinetic Theory of Gases (KTG) and Thermodynamics problems? Here are 15 carefully selected advanced-level questions to help you gear up for JEE Advanced. Make sure you understand the concepts from our KTG and Thermodynamics notes before attempting these questions.

Question 1:

Calculate the number of degrees of freedom for a diatomic gas at room temperature and find the total internal energy of 1 mole of the gas at 300 K. Assume ideal gas behavior.

Solution:

A diatomic gas has 5 degrees of freedom (3 translational and 2 rotational) at room temperature.

Total Internal Energy = \( U = \frac{5}{2} nRT \)

Substituting the values:

\( U = \frac{5}{2} \times 1 \times 8.314 \times 300 = 6235.5 \, \text{J} \)

Hence, the total internal energy is 6235.5 J.

Question 2:

A container with 1 mole of an ideal gas is divided into two chambers. One side is initially evacuated and the other side has the gas at 300 K and 1 atm. The partition is removed, and the gas fills both chambers. Calculate the change in entropy due to this free expansion.

Solution:

The process is isothermal, and for free expansion, the change in entropy can be calculated as:

\( \Delta S = nR \ln \left(\frac{V_f}{V_i}\right) \)

Since the gas fills double the volume, \( V_f = 2V_i \), so:

\( \Delta S = 1 \times 8.314 \times \ln(2) = 5.76 \, \text{J/K} \)

Thus, the change in entropy is 5.76 J/K.

Question 3:

An ideal gas undergoes an adiabatic process. If the pressure of the gas changes from 4 atm to 1 atm and the initial temperature is 400 K, calculate the final temperature of the gas. Assume \( \gamma = 1.4 \).

Solution:

For an adiabatic process:

\( \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} \)

Substituting the values:

\( \frac{T_2}{400} = \left(\frac{1}{4}\right)^{\frac{0.4}{1.4}} \)

\( T_2 = 400 \times (0.25)^{0.2857} = 263.15 \, \text{K} \)

Thus, the final temperature of the gas is approximately 263 K.

Question 15:

A Carnot engine operates between a hot reservoir at 500 K and a cold reservoir at 300 K. If the engine absorbs 1500 J of heat from the hot reservoir in each cycle, calculate the work done by the engine and the heat rejected to the cold reservoir.

Solution:

The efficiency of a Carnot engine is:

\( \eta = 1 - \frac{T_c}{T_h} \)

\( \eta = 1 - \frac{300}{500} = 0.4 \)

The work done by the engine is:

\( W = \eta Q_h = 0.4 \times 1500 = 600 \, \text{J} \)

The heat rejected to the cold reservoir is:

\( Q_c = Q_h - W = 1500 - 600 = 900 \, \text{J} \)

Thus, the engine does 600 J of work and rejects 900 J of heat to the cold reservoir.

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