Pressure remaining constant, at what temperature the volume of a gas will be double of its volume at `0^@C` ?

Options

A. `100^@C`
B. `546^@C`
C. `273 K`
D. `546 K`

Explanation

To solve the problem of determining the temperature at which the volume of a gas will be double its volume at 0°C while keeping the pressure constant, we can use Charles' Law. Here’s a step-by-step solution:

### Step 1: Understand the initial conditions
Let the volume of the gas at 0°C (which is 273 K in Kelvin) be \( V_1 = V \).

### Step 2: Define the final volume
According to the problem, the volume of the gas doubles at some temperature \( T_2 \). Therefore, we have:
\[ V_2 = 2V \]

### Step 3: Apply Charles' Law
Charles' Law states that at constant pressure, the ratio of the volume of a gas to its temperature (in Kelvin) is constant. This can be expressed as:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

### Step 4: Substitute known values
From our definitions:
- \( V_1 = V \)
- \( T_1 = 273 \, K \)
- \( V_2 = 2V \)

Substituting these into Charles' Law gives:
\[ \frac{V}{273} = \frac{2V}{T_2} \]

### Step 5: Simplify the equation
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{1}{273} = \frac{2}{T_2} \]

### Step 6: Solve for \( T_2 \)
Cross-multiplying gives:
\[ T_2 = 2 \times 273 \]
\[ T_2 = 546 \, K \]

### Step 7: Convert to Celsius
To convert the temperature back to Celsius:
\[ T_2 = 546 \, K - 273 = 273 \, °C \]

### Final Answer
The temperature at which the volume of the gas will be double its volume at 0°C is **273°C** or **546 K**.

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