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

Options

A.
B.
C.
D.

Answer

To determine which sketch represents an isobaric process given the condition \((nR)/(P) > 1\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Isobaric Process**:
An isobaric process is one in which the pressure (P) remains constant. Therefore, we can express the relationship between volume (V) and temperature (T) using the ideal gas law:
\[ PV = nRT \]
Rearranging gives:
\[ V = \frac{nRT}{P} \]

2. **Expressing Volume in Terms of Temperature**:
Since P is constant in an isobaric process, we can express the equation as:
\[ V = \frac{nR}{P} \cdot T \]
Here, \(\frac{nR}{P}\) is a constant (let's denote it as \(C\)), so we can rewrite the equation as:
\[ V = CT \]

3. **Logarithmic Form**:
Taking the logarithm of both sides gives:
\[ \log V = \log C + \log T \]
This can be rearranged to:
\[ \log V = \log T + \log C \]
This is in the form of \(y = mx + c\), where \(y = \log V\), \(x = \log T\), \(m = 1\), and \(c = \log C\).

4. **Analyzing the Constant \(C\)**:
Since we are given that \((nR)/(P) > 1\), it implies that \(C > 1\). Therefore, \(\log C > 0\), meaning that the intercept \(c\) is positive.

5. **Graph Characteristics**:
The graph of \(\log V\) versus \(\log T\) will be a straight line with:
- A positive slope of 1 (indicating a direct proportionality).
- A positive y-intercept (since \(c > 0\)).

6. **Identifying the Correct Sketch**:
From the above analysis, we need to find a sketch that shows a straight line starting from a positive value on the y-axis (indicating the positive intercept) and having a slope of 1.

### Conclusion:
The correct sketch representing an isobaric process will be the one that shows a straight line with a positive y-intercept and a slope of 1.