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 \( 2 \times 10^{-5} \, \text{per } ^\circ\text{C} \).

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,
  $C_P > C_V$ 

  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 (
  $W = P\Delta V = 0$ 

  ).

Using the ideal gas equation \(P = \frac{nRT}{V}\), with total moles \(n = 0.2 + 0.3 = 0.5\text{ mol}\) and volume \(V = 200 \times 10^{-6}\text{ m}^3\). Calculating gives \(P = \frac{0.5 \times 8.3 \times 200}{2 \times 10^{-4}} = 4.15 \times 10^6\text{ Pa}\).

Using the relation \(\frac{C}{5} = \frac{F - 32}{9}\), let \(C = F = x\). This gives \(\frac{x}{5} = \frac{x - 32}{9} ⇒ 9x = 5x - 160 ⇒ x = -40\).

Steam has latent heat of vaporization, causing more severe burns than boiling water. Water also has an exceptionally high specific heat capacity compared to most other substances. Both statements are correct.

Heat lost by water to reach \(0^\circ\text{C}\) is \(300 \times 1 \times 25 = 7500\text{ cal}\). Heat required to melt all ice is \(100 \times 80 = 8000\text{ cal}\). Since heat lost is less than heat required to melt all ice, some ice remains, and the final temperature is \(0^\circ\text{C}\).

For isothermal process in A: \(P_A = 2P_0\). For adiabatic process in B: \(P_B = P_0 (2)^\gamma\). The ratio of final pressures is \(\frac{P_B}{P_A} = \frac{P_0 2^\gamma}{2 P_0} = 2^{\gamma - 1}\).

Mean free path \(\lambda \propto V\) and \(v_{\text{rms}} \propto \sqrt{T}\). For an adiabatic process, \(T \propto V^{-(\gamma - 1)}\), so \(v_{\text{rms}} \propto V^{-(\gamma - 1)/2}\). Average time \(\tau = \frac{\lambda}{v_{\text{rms}}} \propto V^{1 + (\gamma - 1)/2} = V^{(\gamma+1)/2}\), so \(q = \frac{\gamma+1}{2}\).

Average translational kinetic energy per molecule is \(\frac{3}{2}kT\), which depends only on temperature and is same for all gases. However, at a given temperature, gas molecules have a distribution of speeds, not the same speed. Assertion is true but Reason is false.

Work and heat depend on the path, whereas internal energy depends only on the initial and final states. For any gas, \(C_P > C_V\) due to expansion work. Since volume is constant, \(dV = 0\), so work done \(W = 0\). Hence, all statements are correct.