According to Charles' law, at constant pressure, 100 ml of a given mass of a gas with `10^@C` rise in temperature will become ( `1/273 ` = 0.00366)

Options

A. 100.0366ml
B. 99.9634ml
C. 103.66ml
D. 100.366ml

Answer

To solve the problem using Charles' Law, we will follow these steps:

### Step 1: Understand Charles' Law
Charles' Law states that the volume of a gas is directly proportional to its absolute temperature (in Kelvin) when pressure is held constant. Mathematically, this can be expressed as:
\[ V \propto T \]
or
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

### Step 2: Identify Given Values
From the question, we have:
- Initial volume, \( V_1 = 100 \, \text{ml} \)
- Initial temperature, \( T_1 = 10^\circ C \)
- Increase in temperature = \( 10^\circ C \)
- Therefore, the final temperature, \( T_2 = T_1 + 10^\circ C = 10^\circ C + 10^\circ C = 20^\circ C \)

### Step 3: Convert Temperatures to Kelvin
To use Charles' Law, we need to convert the temperatures from Celsius to Kelvin:
- \( T_1 = 10^\circ C = 273 + 10 = 283 \, \text{K} \)
- \( T_2 = 20^\circ C = 273 + 20 = 293 \, \text{K} \)

### Step 4: Apply Charles' Law
Now we can apply Charles' Law:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Substituting the known values:
\[ \frac{100 \, \text{ml}}{283 \, \text{K}} = \frac{V_2}{293 \, \text{K}} \]

### Step 5: Solve for \( V_2 \)
Cross-multiplying to solve for \( V_2 \):
\[ V_2 = \frac{100 \, \text{ml} \times 293 \, \text{K}}{283 \, \text{K}} \]

### Step 6: Calculate \( V_2 \)
Now, calculate \( V_2 \):
\[ V_2 = \frac{100 \times 293}{283} \approx 103.53 \, \text{ml} \]

### Conclusion
Thus, the new volume \( V_2 \) after a 10-degree rise in temperature is approximately:
\[ V_2 \approx 103.53 \, \text{ml} \]

### Final Answer
The volume of the gas after a 10-degree rise in temperature will become approximately **103.53 ml**.

---