Question

An ideal gas goes through a reversible isothermal expansion (solid line) followed by a reversible adiabatic expansion (dashed line). Which of the following diagram(s) closely depict(s) the entire process?

• A. (i) and (iii) only
• B. (i) only
• C. (ii) and (iv) only
• D. (i), (ii), and (iii) only

Hint:

Isothermal expansion = Temperature constant, Pressure drops, Volume increases.
Adiabatic expansion = Temperature falls, Pressure drops even faster than in the isothermal step!
Steeper slope on a P-V graph is the signature of an adiabatic curve.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Isothermal Expansion: $T = \text{const}$, $P \downarrow$, $V \uparrow$.
Adiabatic Expansion: $Q = 0$, $P \downarrow$ (more sharply), $V \uparrow$, $T \downarrow$.

Step 2: Detailed Explanation:
$\bullet$ Diagram (i) [P vs V]: Solid line shows $P$ decreasing as $V$ increases. Dashed line (adiabatic) continues the trend but with a steeper slope ($-\gamma P/V$). This is correct.
$\bullet$ Diagram (ii) [T vs V]: Isothermal is a horizontal line ($T = \text{const}$). Adiabatic expansion must show $T$ decreasing as $V$ increases. The graph shows $T$ increasing, which is wrong.
$\bullet$ Diagram (iii) [P vs T]: Isothermal is a vertical line at $T = \text{const}$ (moving down as $P \downarrow$). Adiabatic expansion shows both $P$ and $T$ decreasing together, curving toward the origin. This is correct.
$\bullet$ Diagram (iv) [P vs 1/V]: For isothermal, $P \propto 1/V$ (straight line through origin). The peak shown makes no thermodynamic sense for these processes.

Step 3: Final Answer:
Graphs (i) and (iii) are correct.
This matches option (A).