Question

At 323 K, the vapour pressure in millimeters of mercury of a methanol-ethanol solution is represented by the equation, p = 120 XA + 140, where XA is the mole fraction of methanol. Then the value of $\lim_{x_A \to 1} \frac{p_A}{X_A}$ is (where PA is vapour pressure of A)

• A. 250 mm
• B. 140 mm
• C. 260 mm
• D. 20 mm

Answer 1

Answer: (C)

260 mm

Answer 2

The given equation for the total vapour pressure ($p$) of a methanol-ethanol solution at 323 K is: $p = 120 X_A + 140$ where $X_A$ is the mole fraction of methanol (component A).

For an ideal binary solution, the total vapour pressure is described by Raoult's Law: $p = p_A^* X_A + p_B^* X_B$ where $p_A^*$ and $p_B^*$ are the vapour pressures of pure components A and B, respectively. Since $X_A + X_B = 1$, we have $X_B = 1 - X_A$. Substituting this into Raoult's Law: $p = p_A^* X_A + p_B^* (1 - X_A)$ $p = (p_A^* - p_B^*) X_A + p_B^*$

Comparing this ideal solution equation with the given equation ($p = 120 X_A + 140$): By comparing the constant terms, we identify the vapour pressure of pure component B (ethanol) as: $p_B^* = 140$ mm

By comparing the coefficients of $X_A$, we get: $p_A^* - p_B^* = 120$ mm Substituting the value of $p_B^*$: $p_A^* - 140 = 120$ $p_A^* = 120 + 140 = 260$ mm This is the vapour pressure of pure component A (methanol).

The question asks for the value of $\lim_{X_A \to 1} \frac{p_A}{X_A}$, where $p_A$ is the vapour pressure of A (methanol). For an ideal solution, the partial vapour pressure of component A is given by: $p_A = p_A^* X_A$

Substituting this expression for $p_A$ into the limit: $\lim_{X_A \to 1} \frac{p_A^* X_A}{X_A}$

Since $X_A \to 1$, $X_A$ is not zero, so we can cancel $X_A$: $\lim_{X_A \to 1} p_A^*$

The vapour pressure of pure methanol, $p_A^*$, is a constant value (260 mm). The limit of a constant is the constant itself. Therefore, $\lim_{X_A \to 1} p_A^* = p_A^* = 260$ mm.