Question

1 mole of an ideal gas undergoes the following processes:
Process A \(\rightarrow\) Isothermal expansion at \(400\text{K}\) from volume \(V_1\) to volume \(V_2\), such that \(V_2 = 4V_1\).
Process B \(\rightarrow\) Adiabatic expansion from volume \(V_1\) to volume \(V_2\), such that \(V_2 = 4V_1\).
Consider the following statements and select the correct one/s.

• A. Work done by gas in Process A is greater than in Process B.
• B. Final temperature in Process B is less than 400K.
• C. Change in internal energy is 0 in Process A but non-zero in Process B.
• D. Heat absorbed by the gas is positive in Process A but zero in Process B.

Hint:

For any expansion starting from the same initial state to the same final volume:
- Isothermal expansion always does more work and ends at a higher temperature than adiabatic expansion.
- Always remember that the internal energy of an ideal gas is purely a temperature-dependent state function.

Answer 1

The Correct Option is A, B, C, D

Step 1: Understanding the Concept:
This question compares two paths of expansion for an ideal gas starting from the same initial state. Process A is isothermal ($\Delta T = 0$), and Process B is adiabatic ($q = 0$). We need to evaluate work ($W$), internal energy ($\Delta U$), and heat ($q$) for both.
Step 2: Detailed Explanation:
- Statement (A): Correct. On a $P-V$ diagram, the slope of an adiabatic curve is steeper than an isothermal curve. For the same expansion from $V_1$ to $V_2$, the isothermal curve always stays above the adiabatic curve. Since the area under the curve represents work, $W_{isothermal}>W_{adiabatic}$.
- Statement (B): Correct. In an adiabatic expansion, the gas does work at the expense of its internal energy ($\Delta U = q + W = 0 + W$). Since work done by the gas is positive, internal energy decreases. For an ideal gas, $\Delta U \propto \Delta T$. Therefore, temperature must drop. $T_{final}<400$K.
- Statement (C): Correct. For an ideal gas, internal energy depends only on temperature. In Process A (isothermal), $\Delta T = 0$, so $\Delta U = 0$. In Process B (adiabatic), the temperature changes, so $\Delta U \neq 0$.
- Statement (D): Correct. In Process A, $\Delta U = q + W = 0 \rightarrow q = -W$. Since the gas expands, it does work (work of expansion is negative for $q$ calculation if using $W = -P\Delta V$), thus $q$ is positive (heat is absorbed). In Process B (adiabatic), by definition, $q = 0$.
Step 3: Final Answer:
All four statements (A, B, C, D) are correct descriptions of the thermodynamic processes.