For a given mass of a gas what is the shape of (p) versus `(1/V)` graph at constant temperature ?

Explanation

To determine the shape of the graph of pressure (P) versus \( \frac{1}{V} \) at constant temperature for a given mass of gas, we can follow these steps:

### Step-by-Step Solution:

1. **Start with the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles of gas, \( R \) is the universal gas constant, and \( T \) is the absolute temperature.

2. **Rearranging the Ideal Gas Law**:
To find the relationship between pressure and volume, we can rearrange the equation to express pressure \( P \) in terms of volume \( V \):
\[ P = \frac{nRT}{V} \]

3. **Expressing Pressure in terms of \( \frac{1}{V} \)**:
We can rewrite the equation to express \( P \) as a function of \( \frac{1}{V} \):
\[ P = nRT \cdot \frac{1}{V} \]
Here, \( nRT \) is a constant for a given mass of gas at constant temperature.

4. **Identifying the Relationship**:
From the equation \( P = nRT \cdot \frac{1}{V} \), we can see that pressure \( P \) is directly proportional to \( \frac{1}{V} \). This means that if we plot \( P \) on the y-axis and \( \frac{1}{V} \) on the x-axis, we will get a linear relationship.

5. **Graph Characteristics**:
Since \( P \) is directly proportional to \( \frac{1}{V} \), the graph will be a straight line that passes through the origin. The slope of this line will be equal to \( nRT \).

### Conclusion:
The shape of the graph of pressure (P) versus \( \frac{1}{V} \) at constant temperature for a given mass of gas is a straight line with a positive slope equal to \( nRT \).

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