Which of the following sketches is an isobars (Given : `(nR)/(P) gt 1` )

Options

A.
B.
C.
D.

Answer

To solve the question regarding which of the provided sketches represents isobars, given the condition \( \frac{nR}{P} > 1 \), we can follow these steps:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
From this, we can express the volume \( V \) in terms of pressure \( P \) and temperature \( T \):
\[ V = \frac{nRT}{P} \]

### Step 2: Analyze the Given Condition
We are given the condition:
\[ \frac{nR}{P} > 1 \]
This implies:
\[ V > T \]
This means that for the isobars we are looking for, the volume must be greater than the temperature.

### Step 3: Rewrite the Condition in Logarithmic Form
To find the appropriate graph, we can express the condition in logarithmic terms:
\[ \log V > \log T \]
This means we need to find a graph where the logarithm of volume is greater than the logarithm of temperature.

### Step 4: Evaluate Each Graph
We have four graphs to evaluate:

1. **Graph 1**: \( V = T \) (This means \( \log V = \log T \))
- This does not satisfy our condition since \( \log V \) is not greater than \( \log T \).

2. **Graph 2**: \( V > T \) (This means \( \log V > \log T \))
- This satisfies our condition, making this a valid representation of isobars.

3. **Graph 3**: \( T > V \) (This means \( \log T > \log V \))
- This does not satisfy our condition since \( \log T \) is not greater than \( \log V \).

4. **Graph 4**: Mixed conditions where \( V > T \) at one point and \( T > V \) at another.
- This does not consistently satisfy our condition.

### Conclusion
Based on the evaluation of the graphs, the correct answer is **Graph 2**, where \( \log V > \log T \).