# At 25°C, 3 M of a gas occupy a volume of 4.0 L at 1.9 kPa. If the volume decreases to 2.0 L, what will the new pressure be?

##### 1 Answer

#### Answer:

#### Explanation:

The important thing to realize here is that the *temperature* and *number of moles* of gas are **kept constant**, which should let you know that you can use the equation for **Boyle's Law** to find the new pressure of the gas.

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

In simple terms, when pressure **increases**, volume **decreases**, and when pressure **decreases**, volume **increases**.

This happens because as you *increase* the volume of the gas, you essentially increase the space in which the molecules are moving. Since their *average kinetic speed* is **constant**, i.e. since temperature is constant, the molecules will hit the walls of the container **less frequently** than they did in the *smaller volume* **decrease**.

In your case, the volume of the gas *decreased* to half of its original value, which means that you can expect the pressure to **double**.

Mathematically, this is written as

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

Rearrange the equation to solve for

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

and plug in your values to get

#P_2 = (4.0 color(red)(cancel(color(black)("L"))))/(2.0color(red)(cancel(color(black)("L")))) * "1.9 kPa" = color(green)(|bar(ul(color(white)(a/a)"3.8 kPa"color(white)(a/a)|)))#