The temperature of a given mass of a gas is increased from `19^(@)C` to `20^(@)C` at constant pressure. The volume V of the gas is

Options

A. to `V((20)/(19))`.
B. by `(1)/(273.15)` of its volume at `0^(@)C`.
C. by `(1)/(273.15)` of its volume at 0K.
D. by a factor of `(1)/(273.15)` of its volume at `19^(@)C`.

Answer

To solve the problem, we will use Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature (in Kelvin).

### Step-by-Step Solution:

1. **Convert Celsius to Kelvin**:
- The absolute temperature in Kelvin (K) is calculated by adding 273.15 to the Celsius temperature.
- For 19°C:
\[ T_1 = 19 + 273.15 = 292.15 \, K \]
- For 20°C:
\[ T_2 = 20 + 273.15 = 293.15 \, K \]

2. **Apply Charles's Law**:
- Charles's Law can be expressed as:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
- Where \( V_1 \) is the initial volume, \( V_2 \) is the final volume, \( T_1 \) is the initial temperature, and \( T_2 \) is the final temperature.

3. **Assume Initial Volume**:
- For simplicity, let's assume the initial volume \( V_1 \) is 1 liter (you can choose any value, but it will cancel out).
- Thus, we have:
\[ \frac{1 \, L}{292.15 \, K} = \frac{V_2}{293.15 \, K} \]

4. **Solve for \( V_2 \)**:
- Rearranging the equation to find \( V_2 \):
\[ V_2 = \frac{1 \, L \times 293.15 \, K}{292.15 \, K} \]
- Calculating \( V_2 \):
\[ V_2 \approx 1.0034 \, L \]

5. **Conclusion**:
- The volume of the gas increases from 1 liter to approximately 1.0034 liters when the temperature is increased from 19°C to 20°C at constant pressure.