Question

The given figure represents two isobaric processes for the same mass of an ideal gas, then

• A. \( P_2 \geq P_1 \)
• B. \( P_2 > P_1 \)
• C. \( P_1 = P_2 \)
• D. \( P_1 > P_2 \)

Answer 1

The Correct Option is D

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. 

1. \(P_1\) 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.
2. The slope of the line is inversely proportional to the pressure \( P \), meaning a steeper line indicates lower pressure.
3. 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.

Answer 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:

\[ \text{Slope} \propto \frac{1}{P} \]

Comparing slopes:\[ (\text{Slope})_2 > (\text{Slope})_1 \quad \implies \quad P_2 < P_1 \]