The question involves two isobaric processes for the same mass of an ideal gas as depicted in the graph, where the x-axis represents temperature (\( T \)) and the y-axis represents volume (\( V \)). In isobaric processes, the pressure is constant, but the temperature and volume can change.
From the given graph, the lines \( P_1 \) and \( P_2 \) represent two different isobaric conditions, starting from the origin \( O \). The slope of each line in a \( V-T \) diagram represents the relationship between \( V \) and \( T \) under constant pressure. The steeper the slope, the lower the pressure.
\(P_1\) has a lesser slope compared to \(P_2\). In a \( V-T \) graph for an ideal gas: \(V = \frac{nR}{P}T\) where \( V \) is the volume, \( T \) is the temperature, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( P \) is the pressure.
The slope of the line is inversely proportional to the pressure \( P \), meaning a steeper line indicates lower pressure.
Since \(P_2\) is steeper than \(P_1\), the pressure \( P_2 \) is less than \( P_1 \).
Thus, the correct relationship is \( P_1 > P_2 \), making option \( P_1 > P_2 \) the correct answer.
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Approach Solution -2
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:
