Question

For a perfect gas, two pressures \(P_1\) and \(P_2\) are shown in the figure. The graph shows:

• A. \(P_1>P_2\)
• B. \(P_1<P_2\)
• C. \(P_1 = P_2\)
• D. Insufficient data to draw any conclusion

Answer 1

The Correct Option is A

To determine the relationship between the pressures \(P_1\) and \(P_2\) in the given graph, we need to understand the context of the graph depicting a perfect gas. The graph shows Volume (V) on the y-axis and Temperature (T) on the x-axis for a gas, following the ideal gas law, \(PV = nRT\). 

The lines on the graph represent isobaric (constant pressure) processes. In such a process, the equation \(V/T = \text{constant}\) implies a linear relationship with the slope \(m = V/T\). Therefore:\)

• Steeper slopes indicate a larger volume at the same temperature, hence a lower pressure according to Boyle's Law.
• Shallower slopes indicate a smaller volume at the same temperature, hence a higher pressure.

From the given graph, line \(P_2\) is steeper than line \(P_1\). Thus, line \(P_2\) represents a lower pressure than line \(P_1\), leading to the conclusion:

\(P_1 > P_2\).

Therefore, the correct option is: \(P_1 > P_2\)