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 the ideal gas law and the relationship between volume and temperature at constant pressure.

### Step 1: Understand the Given Information
We know that:
- Mass of gas (m) = 10 g
- Initial volume (V1) = 2 liters
- Initial temperature (T1) at NTP = 273 K (which is the standard temperature)
- Final volume (V2) = 4 liters (since it doubles)

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- P = pressure
- V = volume
- n = number of moles of gas
- R = universal gas constant
- T = temperature in Kelvin

Since the pressure (P) and the number of moles (n) of the gas remain constant, we can use the relationship:
\[ \frac{V1}{T1} = \frac{V2}{T2} \]

### Step 3: Substitute the Known Values
We can substitute the known values into the equation:
- V1 = 2 liters
- T1 = 273 K
- V2 = 4 liters
- T2 = ? (this is what we want to find)

So, the equation becomes:
\[ \frac{2}{273} = \frac{4}{T2} \]

### Step 4: Cross-Multiply to Solve for T2
Cross-multiplying gives us:
\[ 2 \cdot T2 = 4 \cdot 273 \]

### Step 5: Calculate T2
Now, calculate the right side:
\[ 2 \cdot T2 = 1092 \]
Now, divide both sides by 2:
\[ T2 = \frac{1092}{2} = 546 \, \text{K} \]

### Final Answer
Thus, the temperature at which the volume will be double, while keeping pressure and the amount of gas constant, is:
\[ T2 = 546 \, \text{K} \]

---