An ideal gas on heating from 100 K to 109 K shows an increase by `a %` in its volume at constant P. The value of a is _____________.

Explanation

To solve the problem of finding the percentage increase in volume of an ideal gas when heated from 100 K to 109 K at constant pressure, we can follow these steps:

### Step 1: Understand the relationship between volume and temperature
For an ideal gas, at constant pressure, the volume (V) is directly proportional to the temperature (T). This relationship can be expressed as:
\[ V \propto T \]
This means that:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
where \( V_1 \) and \( V_2 \) are the initial and final volumes, and \( T_1 \) and \( T_2 \) are the initial and final temperatures, respectively.

### Step 2: Set up the initial and final conditions
Given:
- Initial temperature, \( T_1 = 100 \, \text{K} \)
- Final temperature, \( T_2 = 109 \, \text{K} \)

Let the initial volume be \( V \). According to the relationship derived in Step 1:
\[ \frac{V}{100} = \frac{V'}{109} \]
where \( V' \) is the final volume.

### Step 3: Solve for the final volume
Rearranging the equation gives:
\[ V' = \frac{109}{100} V \]

### Step 4: Calculate the change in volume
The change in volume, denoted as \( \Delta V \), is given by:
\[ \Delta V = V' - V \]
Substituting the expression for \( V' \):
\[ \Delta V = \frac{109}{100} V - V \]
\[ \Delta V = \left(\frac{109}{100} - 1\right)V \]
\[ \Delta V = \left(\frac{109 - 100}{100}\right)V \]
\[ \Delta V = \frac{9}{100} V \]

### Step 5: Calculate the percentage change in volume
The percentage change in volume is given by:
\[ \text{Percentage Change} = \frac{\Delta V}{V} \times 100 \]
Substituting \( \Delta V \):
\[ \text{Percentage Change} = \frac{\frac{9}{100} V}{V} \times 100 \]
\[ \text{Percentage Change} = \frac{9}{100} \times 100 \]
\[ \text{Percentage Change} = 9\% \]

### Conclusion
Thus, the value of \( a \) is:
\[ a = 9 \]