If the volume of air at `0^(@)C` and `10 atmospheric` pressure is `10 litre`. Its volume, in litre, at normal temperature and pressure would be

Options

A. `1`
B. `10`
C. `100`
D. `1000`

Explanation

To solve the problem, we will use the ideal gas law, which states that \( PV = nRT \). We will apply the combined gas law to find the new volume of air at normal temperature and pressure (NTP).

### Given:
- Initial volume \( V_1 = 10 \) liters
- Initial temperature \( T_1 = 0^\circ C = 273 \, K \)
- Initial pressure \( P_1 = 10 \, atm \)
- Normal temperature \( T_2 = 20^\circ C = 293 \, K \)
- Normal pressure \( P_2 = 1 \, atm \)

### Step 1: Write the combined gas law
The combined gas law relates the initial and final states of a gas:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

### Step 2: Rearrange the equation to find \( V_2 \)
Rearranging the combined gas law to solve for \( V_2 \):
\[ V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1} \]

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:
\[ V_2 = 10 \, L \times \frac{10 \, atm}{1 \, atm} \times \frac{293 \, K}{273 \, K} \]

### Step 4: Calculate \( V_2 \)
Calculating \( V_2 \):
\[ V_2 = 10 \times 10 \times \frac{293}{273} \]
\[ V_2 = 100 \times \frac{293}{273} \]
\[ V_2 \approx 100 \times 1.070 = 107.0 \, L \]

### Step 5: Conclusion
The volume of air at normal temperature and pressure is approximately \( 107.0 \) liters.