Let's analyze and solve the problem step by step: --- ## **Given:** A two-component solution, with pressures of each component: \[ P_1 = x_1 P_1^ \exp(ax_2^2) \] \[ P_2 = x_2 P_2^ \exp(bx_1^2) \] Where: - \( x_1 \) and \( x_2 \) are the mole fractions (\( x_1 + x_2 = 1 \)) - \( P_1^ \), \( P_2^ \) are vapor pressures of pure components - \( a \) and \( b \) are constants **Question:** How are \( a \) and \( b \) related? --- ## **Step 1: Gibbs-Duhem Equation** For a binary solution at constant T and P, the Gibbs-Duhem equation is: \[ x_1 d\ln P_1 + x_2 d\ln P_2 = \] --- ## **Step 2: Calculate \( d\ln P_1 \) and \( d\ln P_2 \) ** ### For \( P_1 \): \[ \ln P_1 = \ln x_1 + \ln P_1^ + a x_2^2 \] \[ d\ln P_1 = \frac{dx_1}{x_1} + 2a x_2 dx_2 \] But since \( x_1 + x_2 = 1 \), \( dx_2 = -dx_1 \): \[ d\ln P_1 = \frac{dx_1}{x_1} + 2a x_2 (-dx_1) = \frac{dx_1}{x_1} - 2a x_2 dx_1 \] ### For \( P_2 \): \[ \ln P_2 = \ln x_2 + \ln P_2^ + b x_1^2 \] \[ d\ln P_2 = \frac{dx_2}{x_2} + 2b x_1 dx_1 \] \[ = \frac{-dx_1}{x_2} + 2b x_1 dx_1 \] --- ## **Step 3: Substitute into Gibbs-Duhem Equation** \[ x_1 d\ln P_1 + x_2 d\ln P_2 = \] Substitute the results: \[ x_1\left(\frac{dx_1}{x_1} - 2a x_2 dx_1\right) + x_2\left( \frac{-dx_1}{x_2} + 2b x_1 dx_1 \right) = \] \[ (dx_1 - 2a x_1 x_2 dx_1) + (-dx_1 + 2b x_1 x_2 dx_1) = \] \[ dx_1 - 2a x_1 x_2 dx_1 - dx_1 + 2b x_1 x_2 dx_1 = \] \[ (-2a x_1 x_2 + 2b x_1 x_2) dx_1 = \] \[ 2(b - a)x_1 x_2 dx_1 = \] --- ## **Step 4: Solution** For all \( x_1 \), \( x_2 \): \[ b - a = \implies a = b \] --- ## **Final Answer** \[ \boxed{a = b} \] **The parameters \( a \) and \( b \) must be equal for the solution to be thermodynamically consistent.**