Thermal Physics - NEET Physics Questions
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Thermal Physics

Question 111: easy

The volume of a metal sphere increases by 0.24% when its temperature is raised by 40ΒΊC. The coefficient of linear expansion of the metal is :

1. \[ 2\times 10^{-5} per ^{oC}\]
2. \[ 6\times 10^{-5} per ^{oC}\]
3. \[ 2.1\times 10^{-5} per ^{oC}\]
4. \[ 1.2\times 10^{-5} per ^{oC}\]
View Answer

The relationship between the coefficient of volume expansion (\( \beta \)) and the coefficient of linear expansion (\( \alpha \)) for a solid is:

\[
\beta = 3\alpha
\]

Given:
- Volume increase = 0.24%
- Temperature increase \( \Delta T = 40^\circ \text{C} \)

The coefficient of volume expansion \( \beta \) is given by:

\[
\beta = \frac{\text{Percentage increase in volume}}{\Delta T} = \frac{0.24}{40} = 0.006\% \, \text{per } ^\circ\text{C} = 6 \times 10^{-5} \, \text{per } ^\circ\text{C}
\]

Now, using \( \beta = 3\alpha \):

\[
\alpha = \frac{\beta}{3} = \frac{6 \times 10^{-5}}{3} = 2 \times 10^{-5} \, \text{per } ^\circ\text{C}
\]

Thus, the coefficient of linear expansion of the metal is \( 2 \times 10^{-5} \, \text{per } ^\circ\text{C} \).

Question 112: moderate

Consider the following statements

A. Work and heat are path functions in thermodynamics.

B. The internal energy of a gaseous system is state function.

C. For gaseous system, CP is greater than CV .

D. Work done by gas at constant volume is zero.

Based on above information pick the correct option.

1. Only statement (A) is correct
2. Only statements (A), (B) and (C) are correct
3. Only statements (B), (C) and (A) are correct
4. All statements (A), (B), (C) and (D) are correct
View Answer

All the given statements are correct:

  • A. Work and heat depend on the path taken during a process, so they are path functions.
  • B. Internal energy depends only on the state of the system, making it a state function.
  • C. For gases,
    CP>CVC_P > C_V
     

    because extra heat is required at constant pressure to do expansion work.

  • D. At constant volume, the gas does no work since volume does not change (
    W=PΞ”V=0W = P\Delta V = 0
     

    ).