Question

The description for a certain brand of house paint claims a coverage of ${}^{460{\mathrm{ft}}^2/\mathrm{gal}}$. (a) Express this quantity in square meters per liter. (b) Express this quantity in an SI unit (see Appendices A and D). (c) What is the inverse of the original quantity, and (d) what is its physical significance?

Short answer

1. $460f{t}^2/gal$is equal to $11.3m^2/L$.
2. $460f{t}^2/gal$is equal to $1.13\times {10}^4{m}^{-1}$
3. The inverse of $460f{t}^2/gal\mathrm{is}2.17\times {10}^{-3}f{t}^2/gal$.
4. From the above obtained reciprocal of the coverage of paint, one can find the thickness of the paint.

Step-by-step solution

Given data

The coverage of paint is $460{\mathrm{ft}}^2\mathrm{per}\mathrm{gallon}$.

Understanding the unit conversion

Chain-link conversions are used for unit conversion in which given values are multiplied successively by conversion factors.

(a) Determination of the quantity in square meters per liter.

1 ft = 0.3048 m

$1\mathrm{gal}=231{\mathrm{inch}}^3\mathrm{and}1{\mathrm{inch}}^3=1.639\times {10}^{-2}\mathrm{L}$

Therefore,

1 gal = 3.79 L

Thus,

$$\begin{gathered} \frac{\left[460\times (0.{3048}^2\right]}{3.79}=11.275 \\ \approx 11.3\frac{{\mathrm{m}}^2}{\mathrm{L}} \end{gathered}$$

Thus, the quantity $460f{t}^2/gal$becomes $11.3{\mathrm{m}}^2/\mathrm{L}$.

(b) Express the quantity in SI units.

The conversion factor is$1m^3=1000L$

Thus, the quantity $460f{t}^2/gal$becomes,

$11.3\frac{m^2}{L}\times \frac{1000L}{1m^3}=1.13\times {10}^4{m}^{-1}$

Thus, the quantity $460f{t}^2/gal$is equal to $1.13\times {10}^4{m}^{-1}$.

(c) Determination of the inverse

The inverse of$460f{t}^2/gal$is calculated as,

$\frac{1}{460f{t}^2/gal}=2.17\times {10}^{-3}gal/f{t}^2$

Thus, the inverse of is $460f{t}^2/gal\mathrm{is}2.17\times {10}^{-3}gal/f{t}^2$.

(d) Physical significance

Since the volume of the paint is to be used to paint the unit area, it is possible to find the thickness of the paint. It is also possible to determine the amount of paint required to paint a certain area with the specified thickness.