SET UP AND SOLVE We can guess at both the mass of gold needed and its density. From these numbers we can then find the volume of gold needed.

Gold sells for around $1500 an ounce. On a particular day, the price might be $1300 or $1700, but this does not significantly affect our assessment. An ounce is about 30 grams. Actually, an ordinary (avoirdupois) ounce is 28.35 g; an ounce of gold is a troy ounce, which is 9.45% greater. Again, this level of error is not important. One hundred dollars worth of gold has a mass somewhere around 2 grams, so a billion (10^9) dollars worth of gold is 20 million (2 \times 10^7) grams, or 20 thousand (2 \times 10^4) kilograms. This corresponds to a weight in British units of around 44,000 lb, or 22 tons.

We can also estimate the volume of this gold. If its density were the same as that of water (1 g/cm^3), the volume would be 2 \times 10^7 cm^3, \text{ or } 20 \ m^3. But gold is a heavy metal; we might guess its density is 10 times that of water. In fact, it is more than 20 times as dense as water. Thus, the volume of gold is about ten times less than the same mass of water. So we expect the volume of gold to be about 2 \ m^3. Visualize two cubical stacks of gold bricks, each 1 meter on a side, and ask whether they would fit in a suitcase!

REFLECT Whether the precise weight is 20 tons or 25 tons doesn’t really matter. Either way, the hero is not about to carry it across the border in a suitcase. And 2 cubic meters certainly won’t fit inside any normal suitcase!

Practice Problem: Suppose our hero needs only a million dollars, not a billion. Are the answers the same or different? Would he do better if he used cut diamonds instead of gold? Answer: The weight is a few hundred pounds; the volume is comparable to a small carry on suitcase.
Diamonds would be better.