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 at constant pressure, the volume of a gas is directly proportional to its absolute temperature (in Kelvin). Mathematically, it can be expressed as:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
where:
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature (in Kelvin)
- \( V_2 \) = final volume
- \( T_2 \) = final temperature (in Kelvin)

### Step 2: Identify the Given Values
From the problem:
- \( V_1 = 100 \, \text{ml} \)
- The initial temperature \( T_1 \) is given as \( 0^\circ C \), which is \( 273 \, \text{K} \).
- The rise in temperature is \( 10^\circ C \), so the final temperature \( T_2 = T_1 + 10 = 273 + 10 = 283 \, \text{K} \).

### Step 3: Set Up the Equation
Using Charles' Law:
\[ \frac{100 \, \text{ml}}{273 \, \text{K}} = \frac{V_2}{283 \, \text{K}} \]

### Step 4: Solve for \( V_2 \)
Cross-multiply to solve for \( V_2 \):
\[ V_2 = 100 \, \text{ml} \times \frac{283 \, \text{K}}{273 \, \text{K}} \]

### Step 5: Calculate \( V_2 \)
Now, calculate:
\[ V_2 = 100 \times \frac{283}{273} \]
To simplify:
\[ \frac{283}{273} \approx 1.0366 \]
Thus,
\[ V_2 \approx 100 \times 1.0366 \approx 103.66 \, \text{ml} \]

### Conclusion
The final volume \( V_2 \) is approximately \( 103.66 \, \text{ml} \).