Match the List-I with List-II

Choose the correct answer from the options given below:
To solve this matching problem, we need to understand the definitions and principles behind each process listed in List-I and find the corresponding process from List-II.
| List-I | List-II |
|---|---|
| A. Pressure varies inversely with volume of an ideal gas. | III. Isothermal process |
| B. Heat absorbed goes partly to increase internal energy and partly to do work. | IV. Isobaric process |
| C. Heat is neither absorbed nor released by a system. | I. Adiabatic process |
| D. No work is done on or by a gas. | II. Isochoric process |
Explanation:
Therefore, the correct answer is A–III, B–IV, C–I, D–II.
Here is the solution:
A - III: Pressure varies inversely with volume of an ideal gas in an isothermal process.
B - IV: If heat is absorbed and it goes to increase internal energy, and partly to do work, it happens in Isobaric process.
C - I: Heat is neither absorbed nor released by the system in Adiabatic process.
D - II: No work is done on or by a gas in Isochoric process.
So, the correct matches are:
A - III
B - IV
C - I
D - II
The effect of temperature on the spontaneity of reactions are represented as: Which of the following is correct?

A gun fires a lead bullet of temperature 300 K into a wooden block. The bullet having melting temperature of 600 K penetrates into the block and melts down. If the total heat required for the process is 625 J, then the mass of the bullet is grams. Given Data: Latent heat of fusion of lead = \(2.5 \times 10^4 \, \text{J kg}^{-1}\) and specific heat capacity of lead = 125 J kg\(^{-1}\) K\(^{-1}\).
Using the given P-V diagram, the work done by an ideal gas along the path ABCD is: 
For a given reaction \( R \rightarrow P \), \( t_{1/2} \) is related to \([A_0]\) as given in the table. Given: \( \log 2 = 0.30 \). Which of the following is true?
| \([A]\) (mol/L) | \(t_{1/2}\) (min) |
|---|---|
| 0.100 | 200 |
| 0.025 | 100 |
A. The order of the reaction is \( \frac{1}{2} \).
B. If \( [A_0] \) is 1 M, then \( t_{1/2} \) is \( 200/\sqrt{10} \) min.
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. \( t_{1/2} \) is 800 min for \( [A_0] = 1.6 \) M.