Question

From the following plots, find the correct option.

• A. \(T_1 \gt T_2 \;;\; P_1 \gt P_2\)
• B. \(T_1 \gt T_2 \;;\; P_2 \gt P_1\)
• C. \(T_2 \gt T_1 \;;\; P_2 \gt P_1\)
• D. \(T_2 \gt T_1 \;;\; P_1 \gt P_2\)

Hint:

In a \(P\)-\(V\) diagram, higher temperature isotherms lie farther from the origin. In a \(V\)-\(T\) graph, slope is inversely proportional to pressure.

Answer 1

The Correct Option is D

Step 1: Analyze the first graph (\(P\) vs \(V\)).
The first graph represents isotherms on a \(P\)-\(V\) diagram.
For an ideal gas, the equation is:
\[ PV = nRT \] At higher temperature, the isotherm lies farther away from the origin because the product \(PV\) becomes larger.
In the graph, the curve labelled \(T_2\) lies above the curve labelled \(T_1\).
Hence,
\[ T_2 \gt T_1 \]

Step 2: Analyze the second graph (\(V\) vs \(T\)).
The second graph represents Charles' law:
\[ V \propto T \] For a fixed amount of gas,
\[ V = \frac{nR}{P}T \] Thus, the slope of the \(V\)-\(T\) graph is:
\[ \text{Slope} = \frac{nR}{P} \] So, slope is inversely proportional to pressure.

Step 3: Compare the pressures.
In the graph, the line corresponding to \(P_2\) has greater slope than the line corresponding to \(P_1\).
Therefore,
\[ P_2 \lt P_1 \] Hence,
\[ P_1 \gt P_2 \]

Step 4: Match with the options.
Combining both results:
\[ T_2 \gt T_1 \quad \text{and} \quad P_1 \gt P_2 \] which corresponds to option (4).

Step 5: Final conclusion.
Therefore, the correct answer is:
\[ \boxed{(4)\; T_2 \gt T_1 \;;\; P_1 \gt P_2} \]