Question

A pressure of `1.0` `atm` will support a `760` `mm` column of mercury. What atmospheric pressure is required to support a column `882` `mm` tall?

Answer 1

A pressure of `1.0` atmosphere (`atm`) supports a column of `760` `mm` of mercury.

Just using proportional reasoning, the pressure we are asked for will be: `P = 882/760 xx 1.0 = 1.16` `atm`

Answer 2

`x=1.16` atm's to 2 decimal places `->` approximate answer

`x=882/760 -> 1 61/32` atm's `->` exact answer

Explanation:

`color(blue)("Preamble")`

`color(brown)(ul("For multiply or divide:"))`
To move a value to the other side of the equals change it into 1. As multiply by 1 does not change the other value.

`color(brown)(ul("For add or subtract:"))`
To move a value to the other side of the equals change it to 0. As adding 0 does not change the other value.

`color(brown)("The shortcut methods use the above but skip steps")`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Answering the question")`

Let the unknown count in atm's be `x`

Known: 1 atm `-=760`mmHg

Using ratio:

`("atm")/("mmHg") ->1/760-=x/882` where `-=` means 'equivalent to

To get `x` on its own multiply both sides by 882.

`color(green)(1/760color(red)(xx882)" "=" "x/882color(red)(xx882))`

`color(green)((1color(red)(xx882))/760" "=" "x xx(color(red)(882))/882 )`

`1.160526...." "=" "x`

`x=1.16` atm's to 2 decimal places `->` approximate answer

`x=882/760 -> 1 61/32` atm's `->` exact answer