Question

The Clausius-Clapeyron equation is: ln (P2/P1)=-(dHvap/R)(1/T2-1/T1). My question is: Why the equation I have seen in a lot of answers has 1/T1 - 1/T2????

Answer 1

Answer:

Both forms of the Clausius-Clapeyron equation are mathematically equivalent—they differ only in sign due to algebraic rearrangement. Here’s why:

1. Standard Form:

$$\ln \left(\frac{P_2}{P_1}\right)=-\frac{\mathrm{\Delta }{H}_{\text{vap}}}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$

• Interpretation:

• The negative sign accounts for the inverse relationship between pressure and temperature.

• Subtracting $\frac{1}{T_2}$ from $\frac{1}{T_1}$ reflects the direction of integration (from $T_1$ to $T_2$).

2. Alternative Form (Flipped Temperatures):

$$\ln \left(\frac{P_2}{P_1}\right)=\frac{\mathrm{\Delta }{H}_{\text{vap}}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$$

• Why It’s Equivalent:

• The terms inside the parentheses are reversed, but the negative sign is absorbed:

  $$-(\frac{1}{T_2}-\frac{1}{T_1})=+(\frac{1}{T_1}-\frac{1}{T_2})$$

• Both forms yield identical results for $\ln (P_2\mathrm{/}{P}_1)$.

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Key Takeaway:

• No physical difference: The choice depends on textbook convention or clarity.

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• Example Calculation:
  If $T_1=300\text{K}$, $T_2=350\text{K}$, and $\mathrm{\Delta }{H}_{\text{vap}}=40\text{kJ/mol}$:

• Both forms give the same $\ln (P_2\mathrm{/}{P}_1)$ value.

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Bottom Line: Use whichever form aligns with your problem’s given variables—the result is the same.