Question

Consider an ideal liquid–vapor mixture at equilibrium having liquid phase mole fraction $(x_i)$ and gas phase mole fraction $(y_i)$ of component $i$. If at a given temperature, $P_{v_i}$ is the vapor pressure of pure component $i$ and $P$ is the total pressure, then the equilibrium ratio $(k_i)$ is

• A. $k_i=\dfrac{x_i}{y_i}=\dfrac{P_{v_i}}{P}$
• B. $k_i=\dfrac{x_i}{y_i}=\dfrac{P}{P_{v_i}}$
• C. $k_i=\dfrac{y_i}{x_i}=\dfrac{P_{v_i}}{P}$
• D. $k_i=\dfrac{y_i}{x_i}=\dfrac{P}{P_{v_i}}$

Hint:

For ideal systems at $T$ = const., combine Raoult’s law ($p_i=x_iP_{v_i}$) with Dalton’s law ($p_i=y_iP$) to get $k_i=\dfrac{y_i}{x_i}=\dfrac{P_{v_i}}{P}$ instantly.

Answer 1

The Correct Option is C

Step 1: Use Raoult’s law for an ideal liquid.
For component $i$, the equilibrium (saturation) partial pressure above the liquid is \[ p_i^{\star} = x_i \, P_{v_i}. \] Step 2: Use Dalton’s law for the gas phase.
At total pressure $P$, the actual partial pressure in the vapor is \[ p_i = y_i \, P. \] Step 3: Equate partial pressures at phase equilibrium.
At equilibrium, \[ p_i = p_i^{\star} \;\Rightarrow\; y_i P = x_i P_{v_i} \;\Rightarrow\; \dfrac{y_i}{x_i} = \dfrac{P_{v_i}}{P}. \] Step 4: Define the equilibrium ratio.
By definition, $k_i \equiv \dfrac{y_i}{x_i} \;\Rightarrow\; k_i = \dfrac{P_{v_i}}{P}$, which matches option (C).