Chapter 2, Problem 2A.1ST

Interpretation Introduction

Interpretation:

The work done when 1.0 mol of propane gas molecules react with oxygen to produce carbon dioxide and liquid water at ${25}^{0}C$ has to be determined.

Concept Introduction:

The force F, opposing an expansion under constant external pressure $p_{ex}$, then $F=p_{ex}A$

The work done is the force when multiplies by the distance moved, is given by

$\begin{array}{l}Workdone,w=Force\times \text{Distance}\\ \\ w=-p_{ex}A\times hwhere,h\times A=\Delta V\\ \\ \therefore w=-p_{ex}\times \Delta V\end{array}$

Explanation of Solution

Change in volume of gas $\Delta V=V_f-V_i$

From Ideal gas equation, $\Delta V=\frac{\Delta nRT}{p_{ex}}$

Here, the atmosphere exerts a constant force on the gas, as the work done by the system loses energy, so it is expected to have a negative value for work done.

$w=-p_{ex}\times \Delta V$

Substituting for $\Delta V$ in the equation for work done as,

$\begin{array}{l}w=-p_{ex}\times \frac{\Delta nRT}{p_{ex}}\\ \\ w=-\Delta nRT\end{array}$

$\Delta n$ is the change in number of moles of only gas molecules is given by $\Delta n=n_f-n_i=3-6=-3$ though number of moles cannot be negative.

$\begin{array}{l}w=-\Delta nRT\\ \\ w=-\left(\left(-3mol\right)\times 8.314J{K}^{-1}mo{l}^{-1}\times \left(25+273K\right)\right)\\ \\ w=+7432.7J=+7.43kJ\end{array}$

The expansion work when dilute hydrochloric acid is poured on to some calcium carbonate is $+7.43kJ$.