Find the approx. number of molecules contained in a vessel of volume `7 `litres at `0^(@)C at 1.3 xx 10^(5)` pascal

Options

A. `2.4 xx 10^23`
B. `3 xx 10^23`
C. `6 xx 10^23 `
D. `4.8 xx 10^23 `

Explanation

To find the approximate number of molecules contained in a vessel of volume 7 liters at 0°C and a pressure of 1.3 x 10^5 pascal, we can use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

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

### Step 1: Convert the volume from liters to cubic meters
1 liter = 0.001 cubic meters, so:
\[ V = 7 \, \text{liters} = 7 \times 0.001 = 0.007 \, \text{m}^3 \]

### Step 2: Convert the temperature from Celsius to Kelvin
To convert Celsius to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]
So:
\[ T = 0 + 273.15 = 273.15 \, \text{K} \]

### Step 3: Rearrange the Ideal Gas Law to solve for \( n \)
We can rearrange the Ideal Gas Law to find the number of moles \( n \):
\[ n = \frac{PV}{RT} \]

### Step 4: Substitute the known values into the equation
Substituting \( P = 1.3 \times 10^5 \, \text{Pa} \), \( V = 0.007 \, \text{m}^3 \), \( R = 8.314 \, \text{J/(mol·K)} \), and \( T = 273.15 \, \text{K} \):
\[ n = \frac{(1.3 \times 10^5) \times (0.007)}{(8.314) \times (273.15)} \]

### Step 5: Calculate \( n \)
Calculating the numerator:
\[ 1.3 \times 10^5 \times 0.007 = 910 \]

Calculating the denominator:
\[ 8.314 \times 273.15 \approx 2270.3 \]

Now, substituting these values:
\[ n = \frac{910}{2270.3} \approx 0.400 \, \text{moles} \]

### Step 6: Calculate the number of molecules
To find the number of molecules, we can use Avogadro's number, which is approximately \( 6.022 \times 10^{23} \, \text{molecules/mol} \):
\[ \text{Number of molecules} = n \times N_A \]
\[ \text{Number of molecules} = 0.400 \times (6.022 \times 10^{23}) \]

### Step 7: Final calculation
Calculating the number of molecules:
\[ \text{Number of molecules} \approx 2.41 \times 10^{23} \]

### Conclusion
The approximate number of molecules contained in the vessel is \( 2.41 \times 10^{23} \).

---