The internal pressure loss of one mole of vander Waal gas over an ideal gas is equal to

Options

A. zero
B. `b^(2)`
C. `(a)/(v^(2)`
D. `b-(a)/(RT)`

Explanation

To solve the problem of determining the internal pressure loss of one mole of a Van der Waals gas compared to an ideal gas, we can follow these steps:

### Step 1: Understand the Ideal Gas Equation
The ideal gas equation is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature

### Step 2: Understand the Van der Waals Equation
The Van der Waals equation for real gases is given by:
\[ \left(P + \frac{n^2A}{V^2}\right)(V - nB) = nRT \]
Where:
- \( A \) and \( B \) are Van der Waals constants specific to the gas.

### Step 3: Identify the Pressure Correction
In the Van der Waals equation, the term \(\frac{n^2A}{V^2}\) represents the pressure correction that accounts for the attractive forces between gas molecules. This means that the actual pressure \( P \) of the Van der Waals gas is less than the pressure predicted by the ideal gas law.

### Step 4: Calculate the Pressure Loss
For one mole of gas (\( n = 1 \)), the pressure loss due to the Van der Waals forces can be expressed as:
\[ \text{Pressure Loss} = \frac{n^2A}{V^2} \]
Substituting \( n = 1 \):
\[ \text{Pressure Loss} = \frac{1^2A}{V^2} = \frac{A}{V^2} \]

### Step 5: Conclusion
Thus, the internal pressure loss of one mole of a Van der Waals gas compared to an ideal gas is:
\[ \text{Internal Pressure Loss} = \frac{A}{V^2} \]

### Final Answer
The internal pressure loss of one mole of Van der Waals gas over an ideal gas is equal to \( \frac{A}{V^2} \).

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