Question:

For van der Waals gases, at the critical point, \( \frac{dP}{dV_m} = 0 \) and

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At the critical point for van der Waals gases, the second derivative of pressure with respect to molar volume is zero, indicating a point of inflection on the pressure-volume curve.
Updated On: Sep 8, 2025
  • \( \frac{d^2 P}{dV_m^2} = 0 \)
  • \( \frac{d^2 P}{dV_m^2}<0 \)
  • \( \frac{d^2 P}{dV_m^2}>0 \)
  • \( \frac{d^2 P}{dV_m^2} \) diverges
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The Correct Option is A

Solution and Explanation

Step 1: Van der Waals equation at the critical point.
For van der Waals gases, the equation of state is: \[ \left( P + \frac{a}{V_m^2} \right) (V_m - b) = RT \] where \( P \) is pressure, \( V_m \) is the molar volume, \( a \) and \( b \) are constants, and \( T \) is the temperature.
At the critical point, the first and second derivatives of pressure with respect to volume become zero, indicating a point of inflection in the pressure-volume curve. This leads to \( \frac{d^2 P}{dV_m^2} = 0 \).
Step 2: Conclusion.
Thus, at the critical point, the second derivative of pressure with respect to molar volume equals zero.
Final Answer: \[ \boxed{\frac{d^2 P}{dV_m^2} = 0} \]
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