Question:
From Clausius-Clapeyron equation, assuming constant $\Delta H_v$, the slope of $\ln P$ vs $1/T$ is:
From Clausius-Clapeyron equation, assuming constant $\Delta H_v$, the slope of $\ln P$ vs $1/T$ is:
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Any equation of form $\ln P = -k(1/T)+C$ has slope $-k$.
Updated On: Jun 29, 2026
- $\frac{\Delta H_v}{R}$
- $-\frac{\Delta H_v}{R}$
- $\frac{R}{\Delta H_v}$
- $-\frac{R}{\Delta H_v}$
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The Correct Option is B
Solution and Explanation
Concept:
Clausius-Clapeyron equation:
\[
\ln P = -\frac{\Delta H_v}{R}\cdot \frac{1}{T} + C
\]
This is of form:
\[
y = mx + c
\]
where slope:
\[
m = -\frac{\Delta H_v}{R}
\]
Step 1: Rewrite equation in linear form.
\[ \ln P = -\frac{\Delta H_v}{R}\left(\frac{1}{T}\right) + C \]
Step 2: Identify slope.
Comparing: \[ \ln P \; \text{vs} \; \frac{1}{T} \] Slope: \[ -\frac{\Delta H_v}{R} \] Final Answer: \[ \boxed{-\frac{\Delta H_v}{R}} \]
Step 1: Rewrite equation in linear form.
\[ \ln P = -\frac{\Delta H_v}{R}\left(\frac{1}{T}\right) + C \]
Step 2: Identify slope.
Comparing: \[ \ln P \; \text{vs} \; \frac{1}{T} \] Slope: \[ -\frac{\Delta H_v}{R} \] Final Answer: \[ \boxed{-\frac{\Delta H_v}{R}} \]
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