Thermal Physics - NEET Physics Questions
← All Chapters

Thermal Physics

Question 201: easy

Assertion (A): When a hot liquid is mixed with a cold liquid, the temperature of the mixer is undefined for some time and then becomes nearly constant.


Reason (R): If two bodies at different temperature are mixed in a calorimeter, the total energy of the two bodies remains conserved.


 

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

When liquids of different temperatures are mixed, it takes time to reach thermal equilibrium, so temperature is not uniform initially. So (A) is true. In an ideal calorimeter, total energy is conserved. So (R) is true. However, (R) does not explain the time-dependent nature of temperature equilibration in (A).

Question 202: easy

Assertion (A): A bottle is filled with water at \(40^{\circ}\text{C}\text{ on opening it at moon, water will boil}\).


Reason (R): Atmospheric pressure on the surface of moon is zero and boiling point is proportional to pressure.


 

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

The moon has a negligible atmosphere, leading to near-zero surface pressure. At such low pressures, water's boiling point decreases significantly. Consequently, water at \(40^{\circ}\text{C}\text{ would readily boil}\). Both (A) and (R) are true, and (R) provides the correct explanation for (A).

Question 203: easy

Assertion (A): The expanded length l of a rod of original length l_0 is not correctly given by assuming \(\alpha\) to be constant with T \( \l = \l_0 (1 + \alpha \Delta T)\), if \(\alpha \Delta T\) is large.


Reason (R): It is given by \(l = \l_0 \text{e}^{\alpha \Delta T}\), which cannot be treated as being approximately equal to \(l_0 (1 + \alpha \Delta T)\text{ for large value of }\alpha \Delta T\).

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

The standard linear expansion formula \(ell = \ell_0 (1 + \alpha \Delta T)\text{ is an approximation valid for small }\alpha \Delta T\), derived from the exponential form \(ell = \ell_0 \text{e}^{\alpha \Delta T}\). If \(alpha \Delta T\) is large, this approximation fails. Both (A) and (R) are true, and (R) correctly explains (A).

Question 204: easy

Assertion (A): Energy of molecules increase on increasing the temperature.


Reason (R): All substances expand on increasing the 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

Assertion (A) is true; average kinetic energy of molecules is directly proportional to absolute temperature. Reason (R) is false because some substances, like water between \(0^{circ}text{C}\) and \(4^{circ}text{C}\,) contract upon heating. Thus, (A) is true and (R) is false.

Question 205: 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 206: 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 207: 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. (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

During free expansion of an ideal gas, no work is done \(W=0\) and no heat is exchanged \(Q=0\). Therefore, change in internal energy \(Delta U = Q - W = 0\). For an ideal gas, \(Delta U = 0\) implies \(Delta T = 0\). Internal energy itself is not zero (it just doesn't change). Free expansion is an irreversible process, so entropy *increases* \(Delta S > 0\), it's not zero. Thus, both (A) and (R) are false.

Question 208: 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. (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 an ideal monatomic gas, molecules only have translational degrees of freedom, and there are no intermolecular forces, so internal energy consists solely of translational kinetic energy. Reason (R) is false; it describes contributions to internal energy from rotational, vibrational, and potential energies which are absent in an *ideal monatomic gas*.

Question 209: 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 210: 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.