Thermal Physics - NEET Physics Questions
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Thermal Physics

Practice Questions:
Question 61: moderate

A solid cube of side \(4\text{ m}\) having coefficient of areal expansion \(2 \times 10^{-5}/^\circ\text{C}\). If temperature is changed by \(40^\circ\text{C}\) then the change in side length of the cube will be

1. \(1.6\text{ mm}\)
2. \(0.16\text{ mm}\)
3. \(3.2\text{ mm}\)
4. \(0.32\text{ mm}\)
View Answer

Coefficient of linear expansion is \(\alpha = \frac{\beta}{2} = 1 \times 10^{-5}/^\circ\text{C}\). Change in length \(\Delta L = L \alpha \Delta T = 4 \times (1 \times 10^{-5}) \times 40 = 1.6 \times 10^{-3}\text{ m} = 1.6\text{ mm}\).

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Question 62: moderate

Thermal capacity of \(40\text{ g}\) of aluminium of specific heat \(0.2\text{ cal/(g }^\{circ}\text{C)}\) is

1. \(40\text{ cal/}^\{circ}\text{C}\)
2. \(160\text{ cal/}^\circ\text{C}\)
3. \(8\text{ cal/}^\circ\text{C}\)
4. \(200\text{ cal/}^\circ\text{C}\)
View Answer

Thermal capacity is given by the formula \(C = m \cdot s\). Substituting the values, \(C = 40\text{ g} \times 0.2\text{ cal/(g }^\circ\text{C)} = 8\text{ cal/}^\circ\text{C}\.

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Question 63: moderate

The equation of state for 14 g nitrogen gas at a pressure P and temperature T, when occupying a volume V will be

1. \(PV = 14RT\)
2. \(PV = \frac{1}{2}RT\)
3. \(PV = RT\)
4. \(PV = 2RT\)
View Answer

Number of moles \(n = \frac{m}{M} = \frac{14\text{ g}}{28\text{ g/mol}} = 0.5\text{ mol}\). Substituting \(n = \frac{1}{2}\) in \(PV = nRT\) gives \(PV = \frac{1}{2}RT\).

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Question 64: moderate
Column I Column II
(A) Total translational kinetic energy (P) $\frac{5}{2} K_B T$
(B) Total rotational kinetic energy (Q) $nRT$
(C) Total kinetic energy per mole (R) $\frac{3}{2} nRT$
(D) Total kinetic energy per molecule (S) $\frac{5}{2} RT$
1. (A)-(R); (B)-(Q); (C)-(S); (D)-(P)
2. (A)-(Q); (B)-(R); (C)-(S); (D)-(P)
3. (A)-(Q); (B)-(R); (C)-(P); (D)-(S)
4. (A)-(R); (B)-(Q); (C)-(P); (D)-(S)
View Answer

Translational KE is \(\frac{3}{2}nRT\). Rotational KE of a rigid diatomic gas is \(nRT\). Total KE per mole is \(\frac{5}{2}RT\) and total KE per molecule is \(\frac{5}{2}K_B T\).

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Question 65: moderate

Consider a sample of \(n\) moles of rigid diatomic gas. Match the columns and tick the correct option (symbols have their usual meanings):

Column I Column II
(A) Total translational kinetic energy (P) $\frac{5}{2}K_B T$
(B) Total rotational kinetic energy (Q) $nRT$
(C) Total kinetic energy per mole (R) $\frac{3}{2}nRT$
(D) Total kinetic energy per molecule (S) $\frac{5}{2}RT$
1. (A)-(R); (B)-(Q); (C)-(P); (D)-(S)
2. (A)-(Q); (B)-(R); (C)-(S); (D)-(P)
3. (A)-(Q); (B)-(R); (C)-(P); (D)-(S)
4. (A)-(R); (B)-(Q); (C)-(S); (D)-(P)
View Answer

Translational KE of \(n\) moles is \(\frac{3}{2}nRT\) (R). Rotational KE is \(nRT\) (Q). KE per mole of diatomic gas is \(\frac{5}{2}RT\) (S). KE per molecule is \(\frac{5}{2}k_B T\) (P). Thus, (A)-(R), (B)-(Q), (C)-(S), (D)-(P).

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Question 66: moderate

Match the columns and tick the correct option. (Symbols have their usual meanings)

\begin{array}{|l|l|}
\hline
\textbf{Column-I} & \textbf{Column-II} \\ \hline
\text{a. } \gamma = \frac{5}{3} & \text{(i) Diatomic gas} \\ \hline
\text{b. } C_v = \frac{5}{2}R & \text{(ii) Triatomic non-linear gas} \\ \hline
\text{c. } C_v = 3R & \text{(iii) Monoatomic gas} \\ \hline
\end{array}

1. a(iii), b(ii), c(i)
2. a(iii), b(i), c(ii)
3. a(i), b(ii), c(iii)
4. a(i), b(iii), c(ii)
View Answer

Monoatomic gas has \(\gamma = 5/3\) (a-iii). Diatomic gas has \(C_v = 5/2 R\) (b-i). Triatomic non-linear gas has \(C_v = 3R\) (c-ii). Thus, the correct matching is a(iii), b(i), c(ii).

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Question 67: moderate

Four moles of helium are mixed with two moles of oxygen. The molar specific heat capacity of the mixture at constant volume is

1. \(\frac{13R}{6}\)
2. \(\frac{11R}{6}\)
3. \(\frac{11R}{2}\)
4. \(\frac{13R}{3}\)
View Answer

For helium (monoatomic), \(C_{v1} = \frac{3}{2}R\) and \(n_1 = 4\). For oxygen (diatomic), \(C_{v2} = \frac{5}{2}R\) and \(n_2 = 2\). The mixture molar specific heat is \(C_{v,\text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{4 \times 1.5R + 2 \times 2.5R}{4 + 2} = \frac{11R}{6}\).

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Question 68: 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.

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Question 69: 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: \(1200 + 15T = 300(50-T) \implies 315T = 13800 \implies T \approx 43.8^\circ\text{C}\).

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Question 70: moderate

Mercury thermometer can be used to measure temperature upto: (1992)

1. $260^\circ\text{C}$
2. $100^\circ\text{C}$
3. $360^\circ\text{C}$
4. $500^\circ\text{C}$
View Answer

The boiling point of mercury is approximately $356.7^\circ\text{C}$. Therefore, a standard mercury thermometer can measure temperatures up to around $360^\circ\text{C}$.

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