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 Question:
We need to compare the thermodynamic parameters (work, temperature, internal energy, and heat) for an isothermal expansion (Process A) and an adiabatic expansion (Process B) of 1 mole of an ideal gas starting from the same initial state ($V_1$, $400\text{K}$) to the same final volume ($V_2 = 4V_1$).


Step 2: Detailed Explanation:
Let us analyze each statement individually:
- Statement (A): On a $P-V$ diagram, the slope of an adiabatic process is steeper than that of an isothermal process ($|(\frac{dP}{dV})_{\text{adi}}| = \gamma |(\frac{dP}{dV})_{\text{iso}}|$ where $\gamma > 1$).
During expansion from $V_1$ to $V_2$, the pressure drops faster in the adiabatic process. Therefore, the isothermal curve lies entirely above the adiabatic curve on a $P-V$ indicator diagram.
Since the area under the curve represents the work done by the gas ($W = \int P \, dV$), the work done in the isothermal expansion (Process A) is greater than in the adiabatic expansion (Process B):
\[ W_{\text{iso}} > W_{\text{adi}} \]
Thus, statement (A) is correct.
- Statement (B): In an adiabatic expansion (Process B), no heat is exchanged with the surroundings ($q = 0$).
According to the first law of thermodynamics:
\[ \Delta U = q + W = 0 - P\Delta V < 0 \]
Since work is done by the gas during expansion ($W_{\text{by gas}} > 0$), the internal energy of the system decreases ($\Delta U < 0$).
For an ideal gas, internal energy is directly proportional to temperature, so a decrease in internal energy leads to a drop in temperature:
\[ T_{\text{final}} < T_{\text{initial}} = 400\text{K} \]
Thus, statement (B) is correct.
- Statement (C): For an ideal gas, the internal energy ($U$) depends only on temperature ($T$).
In Process A (isothermal), temperature is constant ($\Delta T = 0$), so the change in internal energy is zero ($\Delta U_A = 0$).
In Process B (adiabatic), the temperature decreases ($\Delta T \neq 0$), so the change in internal energy is non-zero ($\Delta U_B \neq 0$).
Thus, statement (C) is correct.
- Statement (D):
In Process A (isothermal): $\Delta U = 0 \implies q = W_{\text{by gas}} > 0$. Thus, heat is absorbed by the gas ($q_A > 0$, positive).
In Process B (adiabatic): by definition, no heat exchange occurs, so $q_B = 0$.
Thus, statement (D) is correct.


Step 3: Final Answer:
All statements (A), (B), (C), and (D) are correct.