Question

In the given pressure ($ P $) - absolute temperature ($ T $) graph of an ideal gas, the relation between volumes $ V_1, V_2, V_3, $ and $ V_4 $ is:

• A. \( V_1 = V_2 = V_3 = V_4 \)
• B. \( V_1>V_2>V_3>V_4 \)
• C. \( V_1>V_2>V_3<V_4 \)
• D. \( V_1<V_2<V_3<V_4 \)

Hint:

In a \( P \) vs. \( T \) graph for an ideal gas, the slope of each line is inversely proportional to the volume. Steeper lines correspond to smaller volumes.

Answer 1

The Correct Option is D

We need to determine the relation between volumes \( V_1, V_2, V_3, V_4 \) from a \( P \) vs. \( T \) graph of an ideal gas. Step 1: Use the ideal gas law.
For an ideal gas: \( PV = nRT \). Rewriting: \[ P = \left( \frac{nR}{V} \right) T \] In a \( P \) vs. \( T \) graph, each line has a slope of \( \frac{nR}{V} \), which is inversely proportional to \( V \).
Step 2: Analyze the slopes.
The graph shows \( V_1 \) has the largest slope, and \( V_4 \) the smallest. Since slope \( \propto \frac{1}{V} \): \[ \frac{1}{V_1}>\frac{1}{V_2}>\frac{1}{V_3}>\frac{1}{V_4} \implies V_1<V_2<V_3<V_4 \] Final Answer: \[ \boxed{V_1<V_2<V_3<V_4} \]