Chapter 11, Problem 11.6P

Interpretation Introduction

Interpretation:

Composition of LiCl/H₂O solution at 25⁰C formed adiabatically mixing cool water at 10⁰C with a 20 mol% solution of LiCl/H₂O at 25⁰C should be calculated.

Concept Introduction:

Let us assume $n$ mol of cool water at 10⁰C is mixed adiabatically with 1 mol 20% solution of LiCl/H₂O at 25⁰C. For an adiabatic process, total heat effect is zero.

Following three steps are associated with the preparation of final LiCl/H₂O at 25⁰C.

Enthalpy change for increasing water temperature from 10⁰C to 25⁰C. Let us assume this is denoted as $\Delta H_1$ with unit energy per unit mol. This can be calculated from steam table for water component.

Enthalpy change for preparing 1 mol 20% solution of LiCl/H₂O at 25⁰C is to be calculated. This can be calculated from Figure 11.7 for plot of $\widetilde{\Delta H}$ vs $\tilde{n}$ for LiCl. Here $\tilde{n}$ stands for mol of H₂O per mol of solute. Let us say this is $\widetilde{\Delta H_2}$ . This will supply energy to heat up water from 10⁰C to 25⁰C after cool water is adiabatically mixed up.

Enthalpy change for mixing $\left(n+0\cdot 8\right)$ mol water with 0.2 mol LiCl adiabatically at 25⁰C. As $n$ ( initial mol of cool water) is unknown, trial and error method will be used to determine $\widetilde{\Delta H_3}$ (let us say) using Figure 11.7 for plot of $\widetilde{\Delta H}$ vs $\tilde{n}$ for LiCl till total heat effect is found to be zero. This way, we will get value for $n$ and final composition of LiCl in the solution.

Total heat effect in this entire process is denoted as

  $Q=$ $n$ $\times \Delta H_1$ - 0.2 $\times \widetilde{\Delta H_2}$ +0.2 $\times \widetilde{\Delta H_3}$ .....(1)