Which one of the following represents the graph between log P on Y - axis and 1/T on X - axis ? (P = vapour pressure of a liquid, T = absolute temperature)

Options

A.
B.
C.
D.

Answer

To solve the question regarding the graph between log P (vapor pressure) on the Y-axis and 1/T (inverse of absolute temperature) on the X-axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Clausius-Clapeyron Equation**:
The Clausius-Clapeyron equation describes the relationship between vapor pressure and temperature. It is given by:
\[ \frac{d \ln P}{dT} = \frac{\Delta H_{vap}}{RT^2} \]
where \( P \) is the vapor pressure, \( T \) is the absolute temperature, \( R \) is the universal gas constant, and \( \Delta H_{vap} \) is the enthalpy of vaporization.

2. **Rearranging the Equation**:
We can rearrange the equation to express \( \ln P \) in terms of \( T \):
\[ d \ln P = \frac{\Delta H_{vap}}{RT^2} dT \]

3. **Integrating Both Sides**:
To find the relationship, we integrate both sides. The left side integrates to \( \ln P \), and the right side can be integrated with respect to \( T \):
\[ \ln P = \int \frac{\Delta H_{vap}}{R} \cdot \frac{1}{T^2} dT + C \]
This gives:
\[ \ln P = -\frac{\Delta H_{vap}}{RT} + C \]

4. **Converting to Logarithm Base 10**:
Since we need \( \log P \), we convert \( \ln P \) to \( \log P \):
\[ \log P = \frac{\ln P}{2.303} = -\frac{\Delta H_{vap}}{2.303 R} \cdot \frac{1}{T} + \frac{C}{2.303} \]

5. **Identifying the Linear Relationship**:
The equation can be rearranged to the form of a straight line:
\[ \log P = -\frac{\Delta H_{vap}}{2.303 R} \cdot \frac{1}{T} + \text{constant} \]
Here, \( y = \log P \), \( x = \frac{1}{T} \), and the slope \( m = -\frac{\Delta H_{vap}}{2.303 R} \).

6. **Graph Interpretation**:
The graph of \( \log P \) versus \( \frac{1}{T} \) will be a straight line with a negative slope, indicating that as \( \frac{1}{T} \) increases (or temperature decreases), \( \log P \) decreases.

### Conclusion:
The graph between \( \log P \) on the Y-axis and \( \frac{1}{T} \) on the X-axis is a straight line with a negative slope.