Chapter U5, Problem SIII6RE

(a)

Interpretation Introduction

Interpretation :

The amount of work that the water molecule did on the balloon must be calculated.

Concept Introduction :

Expansion work (W) of a gas can be written as

  ${\text{W=-P}}_{\text{ex}}{\text{(V}}_{\text{2}}{\text{-V}}_{\text{1}}\text{)}\text{.}$

P_ex, V₂ and V₁ are external pressure, final and initial volume respectively.

Answer to Problem SIII6RE

The amount of work that the water molecule did on the balloonis 167.15 J.

Explanation of Solution

  ${\text{W=-P}}_{\text{ex}}{\text{(V}}_{\text{2}}{\text{-V}}_{\text{1}}\text{)}\text{.}$

Putting all the values it can be written as

  $\begin{array}{l}\text{W=-1}\text{.1}\times \text{(1}\text{.6-0}\text{.1)L atm=-1}\text{.1}\times \text{1}\text{.5 L atm=-1}\text{.65 L atm}\\ \text{=-1}\text{.65}\times \text{101}\text{.3 J=-167}\text{.15 J}\text{.}\end{array}$

1 L atm=101.3 J. It is used to get the work in J unit.

b)

Interpretation Introduction

Interpretation :

If there is anything else that needs to be taken into account when calculating the amount of work done by the water vapor must be explained.

Concept Introduction :

For ideal gas work done by the gas ${\text{W=-P}}_{\text{ex}}{\text{(V}}_{\text{2}}{\text{-V}}_{\text{1}}\text{)}\text{.}$

Answer to Problem SIII6RE

The gas inside the balloon must be an ideal gas.

Explanation of Solution

The water vapor transfer heat to the balloon. Then expansion of the gas inside the balloon takes place. Thus work was done by the water vapor to the balloon. This is an expansion work and the formula holds for an ideal gas only. Thus ideality of the gas inside the balloon needs to be taken into account when calculating the amount of work.