If `V_(0)` is the volume of a given mass of gas at 278 K at a constant pressure then according to Charle's law, the volume at `10^(@)C` will be

Options

A. `11 V_(0)`
B. `(1)/(273) (V_(0) + 10)`
C. `V_(0) + (10)/(273)`
D. `(283)/(273) V_(0)`

Explanation

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

### Step 1: Understand Charles's Law
Charles's Law states that the volume of a gas is directly proportional to its temperature (in Kelvin) at constant pressure. Mathematically, it can be expressed as:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

### Step 2: Identify Given Values
From the question, we know:
- \( V_1 = V_0 \) (the initial volume at 278 K)
- \( T_1 = 278 \, \text{K} \) (initial temperature)
- \( T_2 = 10^\circ C \) (the temperature we need to convert to Kelvin)

### Step 3: Convert Celsius to Kelvin
To convert the temperature from Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
Thus,
\[ T_2 = 10 + 273 = 283 \, \text{K} \]

### Step 4: Apply Charles's Law
Now we can apply Charles's Law:
\[ \frac{V_0}{278} = \frac{V_2}{283} \]

### Step 5: Solve for \( V_2 \)
Rearranging the equation to solve for \( V_2 \):
\[ V_2 = V_0 \times \frac{283}{278} \]

### Final Answer
Thus, the volume of the gas at \( 10^\circ C \) will be:
\[ V_2 = V_0 \times \frac{283}{278} \]

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