# 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?

Jun 9, 2017

A pressure of $1.0$ atmosphere ($a t m$) supports a column of $760$ $m m$ of mercury.

Just using proportional reasoning, the pressure we are asked for will be: $P = \frac{882}{760} \times 1.0 = 1.16$ $a t m$

Jun 9, 2017

$x = 1.16$ atm's to 2 decimal places $\to$ approximate answer

$x = \frac{882}{760} \to 1 \frac{61}{32}$ atm's $\to$ exact answer

#### Explanation:

$\textcolor{b l u e}{\text{Preamble}}$

$\textcolor{b r o w n}{\underline{\text{For multiply or divide:}}}$
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.

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To move a value to the other side of the equals change it to 0. As adding 0 does not change the other value.

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$\textcolor{b l u e}{\text{Answering the question}}$

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

Known: 1 atm $\equiv 760$mmHg

Using ratio:

$\left(\text{atm")/("mmHg}\right) \to \frac{1}{760} \equiv \frac{x}{882}$ where $\equiv$ means 'equivalent to

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

$\textcolor{g r e e n}{\frac{1}{760} \textcolor{red}{\times 882} \text{ "=" } \frac{x}{882} \textcolor{red}{\times 882}}$

$\textcolor{g r e e n}{\frac{1 \textcolor{red}{\times 882}}{760} \text{ "=" } x \times \frac{\textcolor{red}{882}}{882}}$

$1.160526 \ldots . \text{ "=" } x$

$x = 1.16$ atm's to 2 decimal places $\to$ approximate answer

$x = \frac{882}{760} \to 1 \frac{61}{32}$ atm's $\to$ exact answer