At what temperature would the volume of a given mass of a gas at constant pressure be twice its volume at `0^(@)C`

Options

A. `546^(@)C`
B. `100^(@)C`
C. `273^(@)C`
D. `373^(@)C`

Answer

To solve the problem of determining the temperature at which the volume of a given mass of gas at constant pressure is twice its volume at \(0^\circ C\), we can follow these steps:

### Step 1: Understand the relationship between volume and temperature
According to Charles's Law, for a given mass of gas at constant pressure, the volume (V) is directly proportional to the absolute temperature (T). This can be expressed as:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

### Step 2: Define the known values
- Let \(V_1\) be the volume of the gas at \(0^\circ C\).
- Therefore, \(V_2\) (the volume at the unknown temperature) is twice \(V_1\), so \(V_2 = 2V_1\).
- The temperature \(T_1\) at \(0^\circ C\) is equivalent to \(273 \, K\).

### Step 3: Substitute the known values into the equation
Now we can substitute the known values into the equation:
\[ \frac{V_1}{T_1} = \frac{2V_1}{T_2} \]

### Step 4: Cancel out the common terms
Since \(V_1\) appears in both terms, we can cancel it out:
\[ \frac{1}{T_1} = \frac{2}{T_2} \]

### Step 5: Rearrange the equation to solve for \(T_2\)
Rearranging gives us:
\[ T_2 = 2 \times T_1 \]

### Step 6: Substitute the value of \(T_1\)
Now substitute \(T_1 = 273 \, K\):
\[ T_2 = 2 \times 273 \, K = 546 \, K \]

### Step 7: Convert the temperature back to Celsius if needed
To convert \(546 \, K\) back to Celsius:
\[ T_{Celsius} = T_{Kelvin} - 273 = 546 - 273 = 273 \, °C \]

### Final Answer
The temperature at which the volume of the gas would be twice its volume at \(0^\circ C\) is \(546 \, K\) or \(273 \, °C\).

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