Thermodynamics - NEET Physics Questions
Question 41: easy

Consider the following statements:


(A) Temperature of a body is related to its average internal energy, not to the kinetic energy of motion of its centre of mass.


(B) The first law of thermodynamics is based on law of conservation of energy applied to any system in which energy transfer from or to the surrounding (through heat and work) is taken into account.


Based on above information, pick correct option.

1. Both statements (A) and (B) are true
2. Both statements (A) and (B) are false
3. Statement (A) is true while (B) is false
4. Statement (B) is true while (A) is false
View Answer

Statement (A) is correct because temperature measures internal, disordered molecular kinetic energy, not ordered bulk kinetic energy. Statement (B) is also correct because the first law of thermodynamics is simply the law of conservation of energy.

Question 42: moderate

The work done by 2 moles of polyatomic gas (\(\gamma = \frac{4}{3}\)) initially at room temperature to increase its volume eight times during adiabatic process will be (Take \(R = 2\text{ cal mol}^{-1}\text{ K}^{-1}\) and room temperature 27°C)

1. 900 cal
2. 600 cal
3. 1800 cal
4. 1200 cal
View Answer

First, find final temperature: \(T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} = 300 \left(\frac{1}{8}\right)^{1/3} = 150\text{ K}\). Then work done \(W = \frac{nR(T_1 - T_2)}{\gamma-1} = \frac{2 \times 2 \times (300 - 150)}{1/3} = 1200\text{ cal}\).

Question 43: easy

Assertion (A): It is possible for both the pressure and volume of a monoatomic ideal gas of a given amount to change simultaneously without causing the internal energy of the gas to change.


Reason (R): The internal energy of an ideal gas of a given amount remains constant if temperature does not change. It is possible to have a process in which pressure and volume are changed such that temperature remains constant.


 

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

For an ideal gas, internal energy \(U\) depends only on temperature \(T\). If \(U\) is constant, then \(T\) is constant. For a constant temperature process (isothermal), pressure \(P\) and volume \(V\) can change while \(T\) remains constant (as \(PV = nRT\)). Thus, both assertion and reason are true, and the reason correctly explains the assertion.

Question 44: 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. (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

Assertion (A) is true. For the same volume expansion, work done \(W = int P dV\). In isothermal expansion, \(P\) drops slower than in adiabatic expansion (due to heat supply), so the area under the \(P-V\) curve is greater for isothermal. Reason (R) is also true and explains why \(P\) behaves differently, leading to different work done.

Question 45: easy

Assertion (A): During the melting of a slab of ice at \(273\text{ K}\) at \(1\text{ atm}\) positive work is done on the ice-water system by the atmosphere.


Reason (R): In above process, the internal energy of ice-water system increases.


 

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

When ice melts to water, its volume decreases \(\Delta V < 0\). Work done *by* the atmosphere *on* the system is \(-P\Delta V\), which is positive. So (A) is true. During melting, latent heat is absorbed, increasing internal energy \(\Delta U = Q - W\). Since \(Q\) is positive and \(W\) (work by system) is negative, \(\Delta U\) is positive. So (R) is true. However, the increase in internal energy is not the reason for the work done by the atmosphere; it's the volume change. So (R) does not explain (A).

Question 46: easy

Assertion (A): It is possible for both the pressure and volume of a monoatomic ideal gas of a given amount to change simultaneously without causing the internal energy of the gas to change.


Reason (R): The internal energy of an ideal gas of a given amount remains constant if temperature does not change. It is possible to have a process in which pressure and volume are changed such that temperature remains constant.


 

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 of an ideal gas depends only on temperature.
Formula: For ideal gas, \( U = f(T) \). For monoatomic, \( U = \frac{3}{2} nRT \). Isothermal process implies \( T \) is constant.
Solution: If \( U \) is constant, then \( T \) is constant. An isothermal process allows simultaneous change in \( P \) and \( V \) while \( T \) (and thus \( U \)) remains constant. Reason correctly explains this.

Question 47: 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
View Answer

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 48: 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 49: 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 50: 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).