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 of how the volume \( V \) of a gas changes when its temperature is increased from \( 19^\circ C \) to \( 20^\circ C \) at constant pressure, we can use Charles's Law. Charles's Law states that the volume of a gas is directly proportional to its absolute temperature (in Kelvin) when the pressure is held constant.

### Step-by-Step Solution:

1. **Convert Celsius to Kelvin**:
- The initial temperature \( T_1 \) in Celsius is \( 19^\circ C \).
- To convert to Kelvin:
\[ T_1 = 19 + 273.15 = 292.15 \, K \]
- The final temperature \( T_2 \) in Celsius is \( 20^\circ C \).
- To convert to Kelvin:
\[ T_2 = 20 + 273.15 = 293.15 \, K \]

2. **Use Charles's Law**:
- According to Charles's Law, the relationship between volume and temperature at constant pressure is given by:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
- Here, \( V_1 \) is the initial volume and \( V_2 \) is the final volume.

3. **Assume Initial Volume**:
- Let’s assume the initial volume \( V_1 \) is \( V_0 \).

4. **Set Up the Equation**:
- Substitute the known values into Charles's Law:
\[ \frac{V_0}{292.15} = \frac{V_2}{293.15} \]

5. **Solve for Final Volume \( V_2 \)**:
- Rearranging the equation to solve for \( V_2 \):
\[ V_2 = V_0 \times \frac{293.15}{292.15} \]

6. **Calculate the Change in Volume**:
- The change in volume can be calculated as:
\[ \Delta V = V_2 - V_0 = V_0 \times \left(\frac{293.15}{292.15} - 1\right) \]

7. **Approximate the Change**:
- Since the change in temperature is small, we can approximate:
\[ \Delta V \approx V_0 \times \left(\frac{1}{292.15}\right) \]

### Final Answer:
The volume \( V \) of the gas increases by approximately \( \frac{V_0}{292.15} \) when the temperature is increased from \( 19^\circ C \) to \( 20^\circ C \) at constant pressure.