# Question #e5462

##### 1 Answer

#### Answer:

#### Explanation:

You are indeed dealing with a gas law problem, a **Boyle's Law** problem to be precise.

According to **Boyle's Law**, when *temperature* and *number of moles* of gas are **kept constant**, volume and pressure have an **inverse relationship**.

When pressure **decreases**, volume *increases*, and when pressure **increases**, volume *decreases*.

Mathematically, Boyle's Law is expressed as

#color(blue)(|bar(ul(color(white)(a/a)P_1V_1 = P_2V_2color(white)(a/a)|)))" "# , where

In your case, you know that a sample of gas occupies a volume of

**Standard pressure** is defined as a pressure of *mmHg* by using the conversion factor

#color(purple)(|bar(ul(color(white)(a/a)color(black)("1 atm " = " 760 torr")color(white)(a/a)|)))#

The final pressure of the gas will thus be

#0.98692color(red)(cancel(color(black)("atm"))) * "760 torr"/(1color(red)(cancel(color(black)("atm")))) = "750.06 torr"#

So, the pressure of the gas decreases from **increase**.

Rearrange the equation to solve for

#P_1V_1 = P_2V_2 implies V_2 = P_1/P_2 * V_1#

Plug in your values to get

#V_2 = (1380 color(red)(cancel(color(black)("torr"))))/(750.06color(red)(cancel(color(black)("torr")))) * "473 mL" = "870.25 mL"#

Rounded to three **sig figs**, the answer will be

#V_2 = color(green)(|bar(ul(color(white)(a/a)"870 mL"color(white)(a/a)|)))#