van der Waal's gas equation can be reduced to virial eqation and virial equation (in terms of volume) is`Z=A+(B)/(V_(m))+(C)/(V_(m)^(2))+……..`
where A =first virial coefficient, B=second virial coefficient ,C = third virial coefficient. The third virial coeffdient of Hg(g) is 625 `(cm^(2)//"mol")^(2)`. What volume is available for movement of 10 moles He(g) atoms present in 50 L vessel?

Options

A. 49.75 L
B. 49.25 L
C. 25 L
D. 50 L

Explanation

To solve the problem, we need to determine the volume available for the movement of 10 moles of helium gas in a 50 L vessel, taking into account the third virial coefficient of mercury.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Third virial coefficient (C) for Hg(g) = 625 cm³/mol².
- Number of moles of He (n) = 10 moles.
- Total volume of the vessel (V) = 50 L.

2. **Convert Units**:
- Since the third virial coefficient is given in cm³/mol², we need to convert the volume of the vessel from liters to cm³:
\[ 50 \, \text{L} = 50 \times 1000 \, \text{cm}^3 = 50000 \, \text{cm}^3 \]

3. **Calculate the Second Virial Coefficient (B)**:
- The third virial coefficient (C) is related to the second virial coefficient (B) as follows:
\[ C = B^2 \]
- Therefore, we can find B:
\[ B = \sqrt{C} = \sqrt{625 \, \text{cm}^3/\text{mol}^2} = 25 \, \text{cm}^3/\text{mol} \]

4. **Calculate the Real Volume Available (V_real)**:
- The real volume available for the gas can be calculated using the formula:
\[ V_{\text{real}} = V_{\text{ideal}} - nB \]
- Substituting the values:
\[ V_{\text{real}} = 50000 \, \text{cm}^3 - (10 \, \text{mol} \times 25 \, \text{cm}^3/\text{mol}) \]
\[ V_{\text{real}} = 50000 \, \text{cm}^3 - 250 \, \text{cm}^3 = 49750 \, \text{cm}^3 \]

5. **Convert Real Volume Back to Liters**:
- Convert the real volume from cm³ back to liters:
\[ V_{\text{real}} = \frac{49750 \, \text{cm}^3}{1000} = 49.75 \, \text{L} \]

### Final Answer:
The volume available for the movement of 10 moles of He(g) atoms present in a 50 L vessel is **49.75 L**.