The given figure represents two isobaric processes for the same mass of an ideal gas, then 
- \( P_2 \geq P_1 \)
- \( P_2 > P_1 \)
- \( P_1 = P_2 \)
- \( P_1 > P_2 \)
The Correct Option is D
Solution and Explanation
From the ideal gas law:
\[ PV = nRT \]
Rearranging for volume:
\[ V = \left( \frac{nR}{P} \right) T \]
The slope of the line in the \( V-T \) graph for an isobaric process is proportional to \( \frac{1}{P} \). Therefore, we have:
\[ \text{Slope} \propto \frac{1}{P} \]
Comparing slopes:\[ (\text{Slope})_2 > (\text{Slope})_1 \quad \implies \quad P_2 < P_1 \]
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