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

### Step-by-Step Solution:

1. **Understand the Concept of Ideal Gas vs. Van der Waals Gas**:
- An ideal gas is a hypothetical gas that perfectly follows the ideal gas law (PV=nRT) without any intermolecular forces or volume occupied by gas molecules.
- A Van der Waals gas accounts for the finite volume of gas molecules and the attractive forces between them.

2. **Identify the Corrections Made by Van der Waals**:
- Van der Waals introduced two main corrections:
- **Volume Correction**: The volume occupied by gas molecules is not negligible. This is represented as \( V_m - b \), where \( b \) is the volume occupied by one mole of gas.
- **Pressure Correction**: The attractive forces between molecules reduce the pressure exerted by the gas. This is represented as \( P + \frac{a}{V_m^2} \), where \( a \) is a constant that accounts for the attraction between molecules.

3. **Write the Van der Waals Equation**:
- The Van der Waals equation is given by:
\[ \left( P + \frac{a}{V_m^2} \right)(V_m - b) = RT \]
- Here, \( P \) is the pressure of the Van der Waals gas, \( V_m \) is the molar volume, \( R \) is the universal gas constant, and \( T \) is the temperature.

4. **Determine the Pressure Loss**:
- The pressure of the ideal gas (\( P_{ideal} \)) can be expressed as:
\[ P_{ideal} = \frac{RT}{V_m} \]
- The pressure of the Van der Waals gas (\( P_{real} \)) can be approximated by rearranging the Van der Waals equation:
\[ P_{real} = \frac{RT}{V_m - b} - \frac{a}{V_m^2} \]
- The internal pressure loss can be calculated as:
\[ \Delta P = P_{ideal} - P_{real} \]

5. **Substituting the Values**:
- Substitute \( P_{ideal} \) and \( P_{real} \) into the equation for pressure loss:
\[ \Delta P = \frac{RT}{V_m} - \left( \frac{RT}{V_m - b} - \frac{a}{V_m^2} \right) \]

6. **Simplifying the Expression**:
- After simplifying, we can express the internal pressure loss in terms of the constants \( a \) and \( b \) and the molar volume \( V_m \).

### Final Result:
The internal pressure loss of one mole of Van der Waals gas over an ideal gas can be expressed as:
\[ \Delta P = \frac{a}{V_m^2} - \frac{RTb}{(V_m)(V_m - b)} \]