# Conversions

Chapter 1R

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The field of dimensional analysis analyzes the relationships between different physical quantities. Refer to significant figures and conversion factors when needed.

## Conversion Factors

A conversion factor is an expression that expresses two equivalent quantities in differing units. They can be written in various ways.

**Inline**

\[1~\mathrm{in} = 2.54~\mathrm{cm}\]

**Fractional**

\[\dfrac{1~\mathrm{in}}{2.54~\mathrm{cm}} \quad \mathrm{or} \quad \dfrac{2.54~\mathrm{cm}}{1~\mathrm{in}}\]

Some conversion factors are *exact* such as the inch-to-centimeter relation given above whereas some are inexact such as the gallon-to-liter relation given as

\[1~\mathrm{gallon~(US)} = 3.785412~\mathrm{L}\]

Here, the quantity for liter has only 7 significant figures.

## Factor-label Method

The **factor-label method** is a technique in algebra used in unit conversion. Conversion factors are expressed as *fractions* and arranged so that any dimensional unit appearing in a numerator and denominator of any of the fractions can be cancelled out until only the desired dimensions remain.

**Example:** Convert 5.2 in to centimeters.

\[5.2~\mathrm{in} \left ( \dfrac{2.54~\mathrm{cm}}{1~\mathrm{in}} \right ) = 1\bar{3}.208~\mathrm{cm} = 13~\mathrm{cm}\]

The final answer has 2 significant figures since the starting value only had 2 significant figures.

## Practice

**Practice**

Assume a lifetime is exactly 80 years. Determine this time frame in

- days
- hours
- minutes

Assume non-leap years.

##
**Solution**

List of needed conversion factors:

1 y = 365 d (exact) 1 d = 24 h (exact)

1 h = 60 min (exact)

1 min = 60 s (exact)

- Convert 80 years to days.

\[80~\mathrm{yr} \left ( \dfrac{365~\mathrm{d}}{1~\mathrm{y}} \right ) = 29,200~\mathrm{d}\]

Note: Because the final answer is exact, there is no ambiguity. We do not have to write it out in scientific notation.

- Convert 80 years to hours

\[80~\mathrm{yr} \left ( \dfrac{365~\mathrm{d}}{1~\mathrm{y}} \right ) \left ( \dfrac{24~\mathrm{h}}{1~\mathrm{d}} \right ) = 716,352~\mathrm{h}\]

Note: Because the final answer is exact, there is no ambiguity. We do not have to write it out in scientific notation.

- Convert 80 years to minutes.

\[ 80~\mathrm{yr} \left ( \dfrac{365~\mathrm{d}}{1~\mathrm{y}} \right ) \left ( \dfrac{24~\mathrm{h}}{1~\mathrm{d}} \right ) \left ( \dfrac{60~\mathrm{min}}{1~\mathrm{h}} \right ) = 42,048,000~\mathrm{min} \]

Note: Because the final answer is exact, there is no ambiguity. We do not have to write it out in scientific notation.

**Practice**

Assume a lifetime is approximately 80.0 years. Determine this time frame in

- days
- hours
- minutes

Assume non-leap years.

##
**Solution**

List of needed conversion factors:

1 yr = 365 d (exact)

1 d = 24 h (exact)

1 h = 60 min (exact)

1 min = 60 s (exact)

- Convert 80 years to days.

\[\begin{align*} 80.0~\mathrm{yr} \left ( \dfrac{365~\mathrm{d}}{1~\mathrm{y}} \right ) &= 29,\bar{2}00~\mathrm{d} \\[1.5ex] &= 2.92\times 10^{4}~\mathrm{d} \end{align*}\]

- Convert 80 years to hours

\[\begin{align*} 80.0~\mathrm{yr} \left ( \dfrac{365~\mathrm{d}}{1~\mathrm{y}} \right ) \left ( \dfrac{24~\mathrm{h}}{1~\mathrm{d}} \right ) &= 70\bar{0},800~\mathrm{h} \\[1.5ex] &= 7.01\times 10^{5}~\mathrm{h} \end{align*}\]

- Convert 80 years to minutes.

\[\begin{align*} 80.0~\mathrm{yr} \left ( \dfrac{365~\mathrm{d}}{1~\mathrm{y}} \right ) \left ( \dfrac{24~\mathrm{h}}{1~\mathrm{d}} \right ) \left ( \dfrac{60~\mathrm{min}}{1~\mathrm{h}} \right ) &= 42,\bar{0}48,000~\mathrm{min}\\[1.5ex] &= 4.20\times 10^{7}~\mathrm{min} \end{align*}\]

**Practice**

A water tank contains exactly 15 gal of water.

- How many 2.0 L bottles would be needed to hold the 15 gal of water?
- How much money (in $) would be needed to purchase the needed number of bottles if each bottle as $0.74?
- If you made $8.25 an hour (after taxes), how many hours (to two decimal places) would you have to work to purchase the needed number of bottles?

##
**Solution**

List of needed conversion factors:

1 gal = 3.785412 L (inexact)

1 bottle = 2.0 L (inexact)

1 bottle = $0.74 (exact)

1 h = $8.25 (exact)

**Needed Bottles**

\[\begin{align*} 15~\mathrm{gal} \left ( \dfrac{3.785412~\mathrm{L}}{1~\mathrm{gal}} \right ) \left ( \dfrac{1~\mathrm{bottle}}{2.0~\mathrm{L}} \right ) &= 28.39059~\mathrm{bottles}\\[1.5ex] &= 29~\mathrm{bottles} \end{align*}\]

Note that we round *this* answer up to the nearest whole number since we are determining the total number of 2.0 L bottles needed to hold the water. 28 of these bottles would be completely filled whereas the 29th bottle would only be about two-fifths of the way full.

**Cost of Bottles**

\[\begin{align*} 29~\mathrm{bottles} \left ( \dfrac{\$0.74}{1~\mathrm{bottle}} \right ) &= \$21.46 \end{align*}\]

Here, 29 bottles represents an exact quantity as well as the cost per bottle.

**Work Hours**

\[\begin{align*} \$21.46 \left ( \dfrac{1~\mathrm{h}}{\$8.25} \right ) &= 2.6012121\ldots \\[1.5ex] &= 2.60~\mathrm{h} \end{align*}\]

Note that this answer is asked to be reported to two decimal places. The quantities involved are all exact numbers the final answer would have at least 11 decimal places which is a bit absurd in this context.

**Practice**

A road trip from New York, NY to Mexico City, Mexico is reported to be 4,145.67 km long.

- How long is the trip in miles?
- Assuming interstate travel with a constant speed of 70.0 mph the entire trip with no stops, how long (in h) would the trip be?
- Suppose the traveler had a lead foot and made the trip at a constant 80.0 mph. How much time (in h) would you save if travelling at this higher speed?

##
**Solution**

List of needed conversion factors:

1 mi = 1.609344 km (exact)

**Length of trip**

\[\begin{align*} 4,145.67~\mathrm{km} \left ( \dfrac{\mathrm{mi}}{1.609344~\mathrm{km}} \right ) &= 2,575.9\bar{9}991052~\mathrm{mi} \\[1.5ex] &= 2,576.00~\mathrm{mi} \end{align*}\]

**Trip duration at 70.0 mph**

Since this is a multi-step calculation, be sure to use the unrounded number from the first calculation in this calculation.

\[\begin{align*} 2,575.999~\mathrm{mi} \left ( \dfrac{\mathrm{h}}{70.0~\mathrm{mi}} \right ) &= 36.\bar{7}999857~\mathrm{h} \\[1.5ex] &= 36.8~\mathrm{h} \end{align*}\]

**Time saved at 80.0 mph**

Since this is a multi-step calculation, be sure to use the unrounded number from the first calculation in this calculation.

\[\begin{align*} 2,575.999~\mathrm{mi} \left ( \dfrac{\mathrm{h}}{80.0~\mathrm{mi}} \right ) &= 32.\bar{1}999875~\mathrm{h} \end{align*}\]

Once again, use the unrounded numbers to obtain an answer and then round off the the appropriate number of significant figures.

\[\begin{align*} 36.\bar{7}999857~\mathrm{h} - 32.\bar{1}999875~\mathrm{h} &= 3.\bar{8}001107~\mathrm{h} \\[1.5ex] &= 3.8~\mathrm{h} \end{align*}\]

Recall, when adding or subtracting, the number of decimal places in the answer is equal to the number of decimal places in the number with the **fewest digits after the decimal.**

**Practice**

What is the heat of vaporization of water (Δ*H*_{vap} in kJ mol^{–1}) if the vapor pressure (*P*) of water is 0.01212 atm at exactly 10 °C (*T*) and 0.46794 atm at at exactly 80 °C?

Use the following equation:

\[\ln \left ( \dfrac{P_2}{P_1} \right ) = \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right )\] Note that the molar gas constant, *R*, is 8.314 J mol^{–1} K^{–1}.

##
**Solution**

Convert *T* into Kelvin.

\[\begin{align*} T_1 = 273.15 + 10~^{\circ}\mathrm{C} &= 283.15~\mathrm{K}~(\text{at}~P_1 = 0.01212~\text{atm})\\[2.0ex] T_2 = 273.15 + 80~^{\circ}\mathrm{C} &= 353.15~\mathrm{K}~(\text{at}~P_2 = 0.46794~\text{atm}) \end{align*}\]

Solve for the heat of vaporization

\[\begin{align*} \ln \left ( \dfrac{P_2}{P_1} \right ) &= \dfrac{\Delta H_{\mathrm{vap}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) \longrightarrow \\[2ex] \Delta H_{\mathrm{vap}} &= R\left (\dfrac{\ln \left ( \dfrac{P_2}{P_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right )\\[2ex] &= (8.314~\mathrm{J~mol^{-1}~K^{-1}}) \left (\dfrac{\ln \left ( \dfrac{0.46794~\mathrm{atm}}{0.01212~\mathrm{atm}} \right )} {\dfrac{1}{283.15~\mathrm{K}} - \dfrac{1}{353.15~\mathrm{K}}}\right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[2ex] &= (8.314~\mathrm{J~mol^{-1}~K^{-1}}) \left (\dfrac{\ln \left ( 38.6\bar{0}8 \right )} {\dfrac{1}{283.15~\mathrm{K}} - \dfrac{1}{353.15~\mathrm{K}}}\right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[2ex] &= (8.314~\mathrm{J~mol^{-1}~K^{-1}}) \left (\dfrac{3.653\bar{4}6} {0.000700039~\mathrm{K^{-1}}}\right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[2ex] &= (8.314~\mathrm{J~mol^{-1}~K^{-1}}) \left ( 5218.\bar{9}7~\mathrm{K}\right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[2ex] &= 43.39~\mathrm{kJ~mol^{-1}} \end{align*}\]