Thermal Physics - NEET Physics Questions
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Thermal Physics

Question 311: easy

The internal energy of an ideal gas depends upon

1. Pressure
2. Temperature
3. Volume
4. Both (1) and (3)
View Answer

The internal energy of an ideal gas is a function of temperature only, as there are no intermolecular forces of attraction in an ideal gas. Therefore, \( U \propto T \).

Question 312: easy

A monoatomic gas does 150 J of work in isothermal expansion. The heat supplied to the gas is

1. 200 J
2. 150 J
3. 100 J
4. Zero
View Answer

For an isothermal process, the change in internal energy is \( \Delta U = 0 \). According to the first law of thermodynamics, \( Q = \Delta U + W \), which gives \( Q = 0 + 150\text{ J} = 150\text{ J} \).

Question 313: moderate

In thermodynamic processes, correct match of column-I with column-II is:

Column-I (Type of process) Column-II (Feature)
a. Isothermal (iv) Temperature constant
b. Isobaric (ii) Pressure constant
c. Isochoric (i) Volume constant
d. Adiabatic (iii) No heat flow between system and surroundings
1. a(i), b(ii), c(iii), d(iv)
2. a(iv), b(i), c(iii), d(ii)
3. a(iv), b(ii), c(iii), d(i)
4. a(iv), b(ii), c(i), d(iii)
View Answer

Isothermal process has constant temperature (a-iv). Isobaric has constant pressure (b-ii). Isochoric has constant volume (c-i). Adiabatic has no heat flow (d-iii). Matching these gives option D.

Question 314: easy

For an ideal gas, total energy is equally distributed in all possible energy modes, with each mode has an average energy equal to \(\frac{1}{2} k_B T\), and each vibrational mode has energy contribution of

1. \(\frac{1}{3} k_B T\)
2. \(k_B T\)
3. \(\frac{3}{2} k_B T\)
4. \(\frac{1}{4} k_B T\)
View Answer

Each vibrational mode has both kinetic energy and potential energy modes, thus having two degrees of freedom. Therefore, the average energy per vibrational mode is \(2 \times \frac{1}{2} k_B T = k_B T\).

Question 315: moderate

\(15\text{ gm}\) of ice at \(0^\circ\text{C}\) is mixed with \(300\text{ gm}\) of water at \(50^\circ\text{C}\) in a container. There is no heat loss due to radiation and water equivalent of container is ignored. What will be final temperature of water?

1. \(5.3^\circ\text{C}\)
2. \(6.7^\circ\text{C}\)
3. \(12.3^\circ\text{C}\)
4. \(43.8^\circ\text{C}\)
View Answer

Heat absorbed to melt ice: \(Q_1 = 15 \times 80 = 1200\text{ cal}\). Let final temperature be \(T\). Heat gained by melted ice: \(15 T\). Heat lost by hot water: \(300(50 - T)\). Equilibrium: \(1200 + 15T = 300(50-T) \implies 315T = 13800 \implies T \approx 43.8^\circ\text{C}\).

Question 316: easy

The relation between coefficient of linear expansion (\(\alpha\)) and coefficient of volume expansion (\(\gamma\)) for solids is

1. \(\gamma = 2\alpha\)
2. \(3\gamma = \alpha\)
3. \(\gamma = 3\alpha\)
4. \(2\gamma = 3\alpha\)
View Answer

Coefficient of volume expansion is three times the coefficient of linear expansion for an isotropic solid, i.e., \(\gamma = 3\alpha\).