An electric bulb of volume `250`cc was sealed during manufacturing at a pressure of `10^(-3)mm` of mercury at `27^(@)C`. Compute the number of air molecules contained in the bulb. Avogadro constant `=6xx10^(23)mol^(-1),` density of mercury`=13600kgm^(-3)` and `g=10ms^(-2)`.

Explanation

To solve the problem of finding the number of air molecules in an electric bulb, we can use the ideal gas law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure in pascals (Pa)
- \( V \) = volume in cubic meters (m³)
- \( n \) = number of moles of gas
- \( R \) = universal gas constant (\( 8.314 \, \text{J/(mol K)} \))
- \( T \) = temperature in kelvins (K)

### Step 1: Convert the given values to appropriate units

1. **Volume**:
\[ V = 250 \, \text{cc} = 250 \times 10^{-6} \, \text{m}^3 = 2.5 \times 10^{-4} \, \text{m}^3 \]

2. **Pressure**:
The pressure is given as \( 10^{-3} \, \text{mm Hg} \). To convert this to pascals, we use the density of mercury and the gravitational acceleration:
\[ P = \text{density} \times g \times h \]
Where:
- Density of mercury \( \rho = 13600 \, \text{kg/m}^3 \)
- \( g = 10 \, \text{m/s}^2 \)
- Height \( h = 10^{-3} \, \text{mm} = 10^{-3} \times 10^{-3} \, \text{m} = 10^{-6} \, \text{m} \)

Therefore,
\[ P = 13600 \times 10 \times 10^{-6} = 0.136 \, \text{Pa} \]

3. **Temperature**:
Convert \( 27^\circ C \) to kelvins:
\[ T = 27 + 273 = 300 \, \text{K} \]

### Step 2: Substitute the values into the ideal gas law

Now we can substitute the values into the ideal gas law equation:
\[ PV = nRT \]
\[ 0.136 \times 2.5 \times 10^{-4} = n \times 8.314 \times 300 \]

### Step 3: Solve for \( n \)

Rearranging the equation to solve for \( n \):
\[ n = \frac{PV}{RT} = \frac{0.136 \times 2.5 \times 10^{-4}}{8.314 \times 300} \]

Calculating the numerator:
\[ 0.136 \times 2.5 \times 10^{-4} = 3.4 \times 10^{-5} \]

Calculating the denominator:
\[ 8.314 \times 300 = 2494.2 \]

Now substituting back:
\[ n = \frac{3.4 \times 10^{-5}}{2494.2} \approx 1.36 \times 10^{-8} \, \text{moles} \]

### Step 4: Convert moles to molecules

To find the number of molecules, we use Avogadro's number:
\[ \text{Number of molecules} = n \times N_A \]
Where \( N_A = 6 \times 10^{23} \, \text{mol}^{-1} \):
\[ \text{Number of molecules} = 1.36 \times 10^{-8} \times 6 \times 10^{23} \]
\[ \text{Number of molecules} \approx 81.6 \times 10^{15} \approx 8.16 \times 10^{16} \]

### Final Answer
The number of air molecules contained in the bulb is approximately \( 8.16 \times 10^{16} \).

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