10 g of a gas at NTP occupies a volume of 2 litres. At what temperature will the volume be double, pressure and amount of the gas remaining same?

Options

A. 273 K
B. 546 K
C. ` –273^(@)C`
D. `546^(@)C`

Explanation

To solve the problem step by step, we will use Charles's Law, which states that for a given mass of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature.

### Step 1: Identify the known values
- Mass of gas (m) = 10 g (not needed for this calculation)
- Initial volume (V1) = 2 liters
- Final volume (V2) = 4 liters (since the volume is doubled)
- Initial temperature (T1) = 0°C = 273 K (at NTP)

### Step 2: Write down Charles's Law
According to Charles's Law:
\[ \frac{V1}{V2} = \frac{T1}{T2} \]
Where:
- \(V1\) = initial volume
- \(V2\) = final volume
- \(T1\) = initial temperature in Kelvin
- \(T2\) = final temperature in Kelvin (which we need to find)

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:
\[ \frac{2 \, \text{liters}}{4 \, \text{liters}} = \frac{273 \, \text{K}}{T2} \]

### Step 4: Simplify the equation
\[ \frac{1}{2} = \frac{273}{T2} \]

### Step 5: Cross-multiply to solve for \(T2\)
Cross-multiplying gives:
\[ T2 = 273 \times 2 \]

### Step 6: Calculate \(T2\)
\[ T2 = 546 \, \text{K} \]

### Conclusion
The final temperature \(T2\) at which the volume will be double, while keeping the pressure and amount of gas constant, is **546 K**.