# Question #35f89

##### 1 Answer

#### Explanation:

You know that, in addition to the *number of moles* of gas present in the sample, the *temperature* of the gas is being **kept constant**.

This should automatically tell you that you're dealing with a situation in which the volume of the gas will depend **exclusively** on its pressure.

When this happens, you can use the equation for **Boyle's Law** to help you find the new volume of the gas

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

What this tells you is that when temperature and number of moles of gas are kept constant, pressure and volume have an **inverse relationship**.

Simply put, when pressure *decreases*, volume **increases**, and when pressure *increases*, volume **decreases**.

In your case, the pressure is **decreasing** from **must increase** as a result of this drop in pressure.

Rearrange the equation to solve for

Plug in your values to get

#V_2 = (0.997color(red)(cancel(color(black)("atm"))))/(0.977color(red)(cancel(color(black)("atm")))) * "5.00 L" = "5.1024 L"#

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

#V_2 = color(green)(|bar(ul(color(white)(a/a)"5.10 L"color(white)(a/a)|)))#

As predicted, the decrease in pressure lead to an increase in volume.