NEET 2016 Phase 1 · ChemistryPrevious Year Question

Consider the following liquid–vapour equilibrium: $Liquid Vapour$ . Which of the following relations is correct?

Answer: (D) $\frac{d ln P}{d T} = \frac{Δ H _{v}}{R T ^{2}}$ . \textbf{Answer:} (D) $\frac{d ln P}{d T} = \frac{Δ H _{v}}{R T ^{2}}$ \textbf{Solution:} This is the Clausius–Clapeyron relation for a liquid–vapour equilibrium.

A. $\frac{d ln G}{d T ^{2}} = \frac{Δ H _{v}}{R T ^{2}}$
B. $\frac{d ln P}{d T} = - \frac{Δ H _{v}}{R T}$
C. $\frac{d ln P}{d T ^{2}} = - \frac{Δ H _{v}}{T ^{2}}$
D. $\frac{d ln P}{d T} = \frac{Δ H _{v}}{R T ^{2}}$ ✓

Correct Answer

(D) $\frac{d ln P}{d T} = \frac{Δ H _{v}}{R T ^{2}}$

Solution & Explanation

\textbf{Answer:} (D) $\frac{d ln P}{d T} = \frac{Δ H _{v}}{R T ^{2}}$ \textbf{Solution:} This is the Clausius–Clapeyron relation for a liquid–vapour equilibrium. Writing $P = K e^{- Δ H_{v} / R T}$ gives $ln P = ln K - \frac{Δ H _{v}}{R T}$ . Differentiating with respect to $T$ (treating $Δ H_{v}$ as constant): $\frac{d ln P}{d T} = \frac{Δ H _{v}}{R T ^{2}}$ .

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Consider the following liquid–vapour equilibrium: Liquid​Vapour. Which of the following relations is correct?

Solution & Explanation

More NEET PYQ solutions

Consider the following liquid–vapour equilibrium: $Liquid Vapour$ . Which of the following relations is correct?