The number of molecules in a litre of a gas at temperature of `27^(@)C` and a pressure of `10^(6)` dyne `cm^(-2)`

Options

A. `2.4xx10^(20)`
B. `2.4xx10^(21)`
C. `2.4xx10^(22)`
D. `2.4xx10^(23)`

Explanation

To find the number of molecules in a litre of a gas at a temperature of \(27^\circ C\) and a pressure of \(10^6 \, \text{dyne/cm}^2\), we can use the ideal gas equation, which is given by:

\[ PV = nRT \]

Where:
- \(P\) = pressure
- \(V\) = volume
- \(n\) = number of moles
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin

However, we want to find the number of molecules (\(N\)), which can be related to the number of moles by:

\[ N = n \cdot N_A \]

Where \(N_A\) is Avogadro's number (\(6.022 \times 10^{23} \, \text{molecules/mol}\)).

### Step 1: Convert the given values to SI units
- **Pressure**: Given as \(10^6 \, \text{dyne/cm}^2\). To convert this to SI units (Pascals):
\[ 1 \, \text{dyne/cm}^2 = 10^{-5} \, \text{N/m}^2 \]
Therefore,
\[ P = 10^6 \, \text{dyne/cm}^2 = 10^6 \times 10^{-5} \, \text{N/m}^2 = 10^1 \, \text{N/m}^2 = 10 \, \text{Pa} \]

- **Volume**: Given as \(1 \, \text{litre}\). To convert this to cubic meters:
\[ V = 1 \, \text{litre} = 1 \times 10^{-3} \, \text{m}^3 \]

- **Temperature**: Given as \(27^\circ C\). To convert this to Kelvin:
\[ T = 27 + 273 = 300 \, \text{K} \]

### Step 2: Use the ideal gas equation
Now we can rearrange the ideal gas equation to solve for \(n\):
\[ n = \frac{PV}{RT} \]

### Step 3: Substitute the values into the equation
Using \(R = 8.314 \, \text{J/(mol K)}\) (which is equivalent to \(8.314 \, \text{m}^3 \cdot \text{Pa}/(\text{mol} \cdot \text{K})\)):
\[ n = \frac{(10 \, \text{Pa})(1 \times 10^{-3} \, \text{m}^3)}{(8.314 \, \text{m}^3 \cdot \text{Pa}/(\text{mol} \cdot \text{K}))(300 \, \text{K})} \]

### Step 4: Calculate \(n\)
Calculating the numerator:
\[ 10 \times 1 \times 10^{-3} = 1 \times 10^{-2} \, \text{Pa m}^3 \]

Calculating the denominator:
\[ 8.314 \times 300 = 2494.2 \, \text{Pa m}^3/\text{mol} \]

Now substituting these values:
\[ n = \frac{1 \times 10^{-2}}{2494.2} \approx 4.01 \times 10^{-6} \, \text{mol} \]

### Step 5: Calculate the number of molecules \(N\)
Now, using \(N = n \cdot N_A\):
\[ N = (4.01 \times 10^{-6} \, \text{mol}) \cdot (6.022 \times 10^{23} \, \text{molecules/mol}) \approx 2.41 \times 10^{18} \, \text{molecules} \]

### Final Answer
The number of molecules in a litre of gas at \(27^\circ C\) and a pressure of \(10^6 \, \text{dyne/cm}^2\) is approximately \(2.41 \times 10^{18}\) molecules.

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