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For a gas obeying the van der Waals' equ...

For a gas obeying the van der Waals' equation, at the critical point

A

both `((delP)/(delV))_(T)`and `((del^(2)P)/(delV^(2)))_(T)`are zero

B

only `((del^(2)P)/(delV^(2)))_(T)` is not zero

C

`((delP)/(delV))_(T)` is zero but `((del^(2)P)/(delV^(2)))_(T)` is non-zero

D

`((delP)/(delV))_(T)` is non-zero but `((del^(2)P)/(delV^(2)))_(T)` is zero

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To solve the question regarding the conditions for a gas obeying the van der Waals equation at the critical point, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Van der Waals Equation**: The van der Waals equation for one mole of a gas is given by: \[ \left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT \] where \( P \) is the pressure, \( V_m \) is the molar volume, \( R \) is the universal gas constant, \( T \) is the temperature, and \( a \) and \( b \) are constants specific to the gas. 2. **Identify the Critical Point**: At the critical point, the gas exhibits unique properties where the distinction between liquid and gas phases disappears. This is characterized by specific conditions of temperature, pressure, and volume. 3. **Derivatives at the Critical Point**: At the critical point, the first and second derivatives of pressure with respect to volume at constant temperature must be equal to zero: \[ \left(\frac{\partial P}{\partial V}\right)_T = 0 \] \[ \left(\frac{\partial^2 P}{\partial V^2}\right)_T = 0 \] This indicates that the slope of the pressure-volume curve is flat (horizontal) at the critical point. 4. **Implication of the Derivatives**: The first derivative being zero indicates that there is no change in pressure with a change in volume at the critical point, while the second derivative being zero indicates that the curvature of the pressure-volume graph is also flat at this point. 5. **Conclusion**: Therefore, for a gas obeying the van der Waals equation at the critical point, both the first and second derivatives of pressure with respect to volume at constant temperature are zero. ### Final Answer: At the critical point for a gas obeying the van der Waals equation: \[ \left(\frac{\partial P}{\partial V}\right)_T = 0 \quad \text{and} \quad \left(\frac{\partial^2 P}{\partial V^2}\right)_T = 0 \]

To solve the question regarding the conditions for a gas obeying the van der Waals equation at the critical point, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Van der Waals Equation**: The van der Waals equation for one mole of a gas is given by: \[ \left(P + \frac{a}{V_m^2}\right)(V_m - b) = RT ...
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Knowledge Check

  • In the van der Waals equation

    A
    `b` is the volume occupied by the gas molecules
    B
    `b` is four times the volume occupied by the gas molecules
    C
    `b` is the correction factor for intermolecular attraction
    D
    None of these
  • van der Waal's equation is true for

    A
    Ideal gas
    B
    Real gas
    C
    Gaseous substance
    D
    None of the above
  • A gas described by van der Waals equation

    A
    behaves similar to an ideal gas in the limit gas in the limit of large molar volumes
    B
    behaves similar to an ideal gas in the limit of large pressures
    C
    is characterised by van der Waals coefficients that are dependent on identity of the gas but are independent of the pressure
    D
    has the pressure that is lower than the pressure exerted by the same behaving ideally
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