If temperature and volume are same, the pressure of a gas obeying van der Waal's equation is :

Options

A. less than that of an ideal gas
B. more than that of an ideal gas
C. same as that of an ideal gas
D. none of these

Explanation

To solve the problem, we need to analyze the Van der Waals equation for real gases and derive the expression for pressure when temperature and volume are constant.

### Step-by-Step Solution:

1. **Understand the Van der Waals Equation**:
The Van der Waals equation for one mole of a real gas is given by:
\[ \left(P + \frac{a}{V^2}\right)(V - b) = RT \]
where \(P\) is the pressure, \(V\) is the volume, \(T\) is the temperature, \(R\) is the universal gas constant, and \(a\) and \(b\) are constants specific to the gas.

2. **Rearranging the Equation**:
We can rearrange the Van der Waals equation to isolate \(P\):
\[ P + \frac{a}{V^2} = \frac{RT}{V - b} \]
\[ P = \frac{RT}{V - b} - \frac{a}{V^2} \]

3. **Substituting Constant Values**:
Since the problem states that temperature \(T\) and volume \(V\) are constant, we can denote them as constants. Thus, we can express \(P\) in terms of these constants:
\[ P = \frac{RT}{V - b} - \frac{a}{V^2} \]

4. **Analyzing the Expression**:
The first term \(\frac{RT}{V - b}\) represents the pressure contribution from the ideal gas behavior, while the second term \(-\frac{a}{V^2}\) accounts for the intermolecular forces (attraction) present in real gases.

5. **Conclusion on Pressure**:
The pressure \(P\) calculated from the Van der Waals equation will be less than the pressure predicted by the ideal gas law because of the negative term \(-\frac{a}{V^2}\). Therefore, under the same temperature and volume, the pressure of a gas obeying the Van der Waals equation is lower than that of an ideal gas.

### Final Answer:
The pressure of a gas obeying Van der Waals equation, when temperature and volume are the same, is less than the pressure predicted by the ideal gas law.

---