A sample of gas has a volume of `V_(1)` litre at temperature `t_(1).^(@)C`. When the temperature of the gas is changed to `t_(2).^(@)C` at constant pressure, then the volume of the gas was found to increase by 10%. The percentage increase in temperature is

Options

A. 0.1
B. `(10+(2730)/(t_(1)))%`
C. `20%`
D. `(0.1+t_(1)^(-1))%`

Answer

To solve the problem step by step, we will use the ideal gas law principles, particularly focusing on the relationship between volume and temperature at constant pressure.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The initial volume of the gas is \( V_1 \) liters at a temperature of \( t_1 \) °C.
- Convert \( t_1 \) to Kelvin:
\[ T_1 = t_1 + 273 \]

2. **Volume Change**:
- The volume increases by 10%. Therefore, the new volume \( V_2 \) can be calculated as:
\[ V_2 = V_1 + 0.10 \times V_1 = 1.10 \times V_1 \]

3. **Using Charles's Law**:
- According to Charles's Law, at constant pressure, the ratio of volume to temperature is constant:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
- Rearranging gives:
\[ \frac{V_1}{1.10 \times V_1} = \frac{T_1}{T_2} \]
- Simplifying this, we find:
\[ \frac{1}{1.10} = \frac{T_1}{T_2} \]

4. **Cross Multiplying**:
- Cross-multiplying gives:
\[ T_2 = 1.10 \times T_1 \]

5. **Calculating the Increase in Temperature**:
- The increase in temperature can be calculated as:
\[ \Delta T = T_2 - T_1 = 1.10 \times T_1 - T_1 = 0.10 \times T_1 \]

6. **Finding the Percentage Increase in Temperature**:
- The percentage increase in temperature is given by:
\[ \text{Percentage Increase} = \left( \frac{\Delta T}{T_1} \right) \times 100 = \left( \frac{0.10 \times T_1}{T_1} \right) \times 100 = 10\% \]

### Final Result:
The percentage increase in temperature is **10%**.