Chapter 3, Problem 3.67P

Interpretation Introduction

(a)

Interpretation:

The given equation should be converted to one relating P(psig), V(ft³), n (lb-mole) and T (°F).

$\text{PV=0}\text{.08206nT}$

Concept introduction:

The equation given is in standard unit as $\text{PV=0}\text{.08206nT}$.

Here, it has to be converted into FPS unit for which different equality factors and conversations have to be used.

Interpretation Introduction

(b)

Interpretation:

The total amount of gas (lb-mole) and the mass of CO (lb_m) in the tank should be calculated.

Concept introduction:

The derived equation must be used as:

$\begin{array}{l}{\text{(P}}^{\text{'}}\text{+14}{\text{.696)×V}}^{\text{'}}\text{=10}{\text{.73n}}^{\text{'}}{\text{×(T}}^{\text{'}}\text{+459}\text{.4)}\\ {\text{n}}^{\text{'}}\text{=}\frac{{\text{(P}}^{\text{'}}\text{+14}{\text{.696)×V}}^{\text{'}}}{\text{10}{\text{.73×(T}}^{\text{'}}\text{+459}\text{.4)}}\end{array}$

Interpretation Introduction

(c)

Interpretation:

The temperature (°F) which is required to gain certain gas pressure of 300 psig, the rated safety limit of cylinder should be calculated.

Concept introduction:

The derived equation should be used as:

$\begin{array}{l}{\text{(P}}^{\text{'}}\text{+14}{\text{.696)×V}}^{\text{'}}\text{=10}{\text{.73n}}^{\text{'}}{\text{×(T}}^{\text{'}}\text{+459}\text{.4)}\\ {\text{T}}^{\text{'}}\text{=}\frac{{\text{(P}}^{\text{'}}\text{+14}{\text{.696)×V}}^{\text{'}}}{\text{10}{\text{.73×n}}^{\text{'}}}\text{-459}\text{.4}\end{array}$