Thermal Physics - NEET Physics Questions
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Thermal Physics

Question 61: easy

Assertion (A): Work done by a gas in isothermal expansion is more than the work done by the gas in the same expansion adiabatically.


Reason (R): Temperature remains constant in isothermal expansion but not in adiabatic expansion.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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Concept: Work done is area under P-V curve. Isothermal vs. Adiabatic expansion.
Formula: \( W = \int P dV \). Isothermal: \( PV = \text{constant} \). Adiabatic: \( PV^{\gamma} = \text{constant} \), where \( \gamma > 1 \).
Solution: During expansion from the same initial state to the same final volume, the pressure in isothermal process drops slower than in adiabatic process, leading to more work done. Temperature remains constant in isothermal and changes in adiabatic.

Question 62: easy

Assertion (A): During free expansion of an Ideal gas, entropy is zero.


Reason (R): Internal energy of an ideal gas is zero during free expansion.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
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Concept: Free expansion of an ideal gas. Entropy change. Internal energy.
Formula: For ideal gas, \( \Delta U = 0 \) (as \( Q=0, W=0 \)). Entropy change \( \Delta S > 0 \) for irreversible free expansion.
Solution: During free expansion of an ideal gas, \( \Delta U = 0 \) (meaning \( T \) is constant), but the internal energy itself is not zero. Also, free expansion is irreversible, so entropy *increases* (not zero). Both A and R are false.

Question 63: easy

Assertion (A): In an ideal monoatomic gas, The Internal energy of gas is equal to translational Kinetic energy of all its molecules


Reason (R): The Internal energy may get contributes from Translational, Rotatory, vibrationally as well as from the Potential energy corresponding to the molecular force.


 

1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer

Concept: Internal energy components for different types of gases.
Formula: For monoatomic ideal gas, \( U = \frac{3}{2} nRT \) (translational only).
Solution: For an ideal monoatomic gas, internal energy is purely translational kinetic energy. General internal energy can have translational, rotational, vibrational, and potential contributions (for real/complex gases), but potential energy is zero for ideal gases. A is true, R is true but not an explanation for A.

Question 64: easy

Assertion (A): Bursting of balloon is not a equilibrium state.


Reason (R): Equilibrium state of a thermodynamic system is completely described by specific values of some macroscopic properties.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Concept: Thermodynamic equilibrium. A bursting balloon is a spontaneous, non-equilibrium process. An equilibrium state is characterized by constant macroscopic properties. Both assertion (A) and reason (R) are true, and (R) correctly explains (A).

Question 65: easy

Assertion (A): Work and heat both can be converted into each other in any condition.


Reason (R): Work and Heat both are different form of energy.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Concept: First Law of Thermodynamics and nature of work/heat. Work and heat are forms of energy transfer, not different forms of energy itself. Their interconversion is governed by thermodynamic laws and not possible under 'any condition'. Both (A) and (R) are false.

Question 66: easy

Assertion (A): If volume of a gas is increasing but temperature of the gas is decreasing, then heat given to the gas may be positive, negative or zero.


Reason (R): Heat given to a gas is a path function, it is not a state function.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Concept: First Law of Thermodynamics (( Delta U = Q - W )). If volume increases, (W > 0) (work done by gas). If temperature decreases, ( Delta U < 0 ). So, ( Q = Delta U + W ) can be positive, negative, or zero. Heat is indeed a path function. Both (A) and (R) are true, and (R) explains (A).

Question 67: easy

Assertion (A): Molar heat capacity of a gas in any process can have any value ( -infty ) to ( +infty ).


Reason (R): Molar heat capacity of a gas in an isothermal process is ( infty ).


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Concept: Molar heat capacity \( C = \frac{dQ}{dT} \). For an isothermal process, ( dT = 0 ), so ( C = infty ). For an adiabatic process, ( dQ = 0 ), so ( C = 0 ). Thus, molar heat capacity can range from ( -infty ) to ( +infty ). Both (A) and (R) are true, and (R) explains (A).

Question 68: easy

Assertion (A): The area of entropy versus temperature graph of a cyclic process, is equal to work done.


Reason (R): Change in internal energy of cyclic process is zero.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Concept: T-S diagram and cyclic processes. The area enclosed by a (T-S) diagram for a cyclic process represents the net heat exchanged, ( Q_{net} ), not the work done. For a cyclic process, ( Delta U = 0 ) is true, but Assertion (A) is false. Therefore, both (A) and (R) are false in relation to the explanation.

Question 69: easy

Assertion (A): Absolute zero temperature is not the zero energy temperature.


Reason (R): At absolute zero temperature the gas may possess potential energy.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Concept: Absolute zero and molecular energy. At absolute zero 0 K, the average kinetic energy of molecules is at its minimum (or zero for an ideal gas classically). However, molecules can still possess potential energy due to intermolecular forces or external fields. Hence, total energy is not necessarily zero. Both (A) and (R) are true, and (R) correctly explains (A).

Question 70: easy

Assertion (A): For gas molecules absolute zero temperature is not the temperature of zero energy.


Reason (R): Only the kinetic energy of the molecules is represented by temperature.


 

1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer

Concept: Temperature and molecular energy. Temperature is a direct measure of the average translational kinetic energy of molecules. While kinetic energy is minimal at 0 K, gas molecules can still have potential energy from intermolecular interactions. Thus,0 K is not zero total energy. Both (A) and (R) are true, and (R) explains (A).