Question:

The volume correction factor for a non-ideal gas in terms of critical pressure ($p_c$), critical molar volume ($V_c$), critical temperature ($T_c$), and gas constant ($R$) is

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For van der Waals gases: $a = \dfrac{27R^2T_c^2}{64p_c}$ and $b = \dfrac{RT_c}{8p_c}$. These relations connect real gas constants with critical constants.
Updated On: Dec 5, 2025
  • $\dfrac{RT_c}{8p_c}$
  • $\dfrac{27R^2T_c^2}{64p_c}$
  • $\dfrac{8p_cV_c}{3T_c}$
  • $3p_cV_c^2$
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The Correct Option is A

Solution and Explanation

Step 1: Start with van der Waals constants.
For a real gas, \[ (p + \frac{a}{V_m^2})(V_m - b) = RT \] At the critical point: \[ p_c = \frac{a}{27b^2}, \quad V_c = 3b, \quad T_c = \frac{8a}{27Rb} \] Step 2: Express $b$ (the volume correction factor) in terms of $R$, $T_c$, and $p_c$.
From the $T_c$ expression: \[ a = \frac{27RbT_c}{8} \] Substitute into the $p_c$ expression: \[ p_c = \frac{27RbT_c}{8} \times \frac{1}{27b^2} = \frac{RT_c}{8b} \] \[ \Rightarrow b = \frac{RT_c}{8p_c} \] Step 3: Conclusion.
The volume correction factor $b = \dfrac{RT_c}{8p_c}$.
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