A chamber containing a gas was evacuated till the vacuum attained was `10^(-14)` m of Hg. If the temperature of the chamber was `30^(@)C,` the number of molecules that remains in it per cubic metre is

Options

A. `3.2xx10^(11)`
B. `3.2xx10^(12)`
C. `2.3xx10^(12)`
D. `2.3xx10^(10)`

Explanation

To solve the problem of finding the number of gas molecules remaining in a chamber evacuated to a vacuum of \(10^{-14}\) m of Hg at a temperature of \(30^\circ C\), we can follow these steps:

### Step 1: Convert the vacuum pressure from mm of Hg to Pascals
The pressure \(P\) in Pascals can be calculated using the formula:
\[ P = \text{density of mercury} \times g \times h \]
where:
- Density of mercury (\(\rho\)) = \(13600 \, \text{kg/m}^3\)
- Gravitational acceleration (\(g\)) = \(10 \, \text{m/s}^2\)
- Height (\(h\)) = \(10^{-14} \, \text{m}\)

Calculating \(P\):
\[ P = 13600 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times 10^{-14} \, \text{m} = 1.36 \times 10^{-10} \, \text{Pa} \]

### Step 2: Convert the temperature from Celsius to Kelvin
To convert the temperature from Celsius to Kelvin:
\[ T = 30^\circ C + 273.15 = 303.15 \, K \]

### Step 3: Use the ideal gas law to find the number of moles
The ideal gas law is given by:
\[ PV = nRT \]
where:
- \(P\) = pressure in Pascals
- \(V\) = volume in cubic meters (1 m³)
- \(n\) = number of moles
- \(R\) = universal gas constant = \(8.31 \, \text{J/(mol K)}\)
- \(T\) = temperature in Kelvin

Rearranging for \(n\):
\[ n = \frac{PV}{RT} \]
Substituting the values:
\[ n = \frac{(1.36 \times 10^{-10} \, \text{Pa})(1 \, \text{m}^3)}{(8.31 \, \text{J/(mol K)})(303.15 \, K)} \]

Calculating \(n\):
\[ n = \frac{1.36 \times 10^{-10}}{8.31 \times 303.15} \approx 5.4 \times 10^{-12} \, \text{mol} \]

### Step 4: Calculate the number of molecules
The number of molecules \(N\) can be calculated using Avogadro's number (\(N_A = 6.02 \times 10^{23} \, \text{molecules/mol}\)):
\[ N = n \times N_A \]
Substituting the values:
\[ N = (5.4 \times 10^{-12} \, \text{mol}) \times (6.02 \times 10^{23} \, \text{molecules/mol}) \approx 3.25 \times 10^{12} \, \text{molecules} \]

### Final Answer
The number of molecules that remain in the chamber per cubic meter is approximately \(3.2 \times 10^{12}\).