# The pressure of 4.2 L of nitrogen gas in a flexible container is decreased to one-half its original pressure, and its absolute temperature is increased to double the original temperature. What is its volume now?

##### 1 Answer

#### Explanation:

Your tool of choice for this problem will be the combined gas law equation, which looks like this

#color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "# , where

Now, even without doing any calculations, you can examine the above equation and predict what you expect the volume of the gas to be at the final state.

Notice that *pressure* and *volume* have an **inverse relationship** when *temperature* **is kept constant** - this is known as Boyle's Law.

This tells you that a **decrease** in pressure will cause an **increase** in volume.

On the other hand, *temperature* and *volume* have a **direct relationship** when *pressure* is **kept constant** - this is known as Charles' Law.

This tells you that an **increase** in temperature will cause an **increase** in volume.

In your case, these two effects will actually **combine**. Increasing the temperature of the gas **while** decreasing its pressure will result in an **increase** in volume, regardless of the actual changes in temperature and pressure.

Mathematically, you can prove this by rearranging the combined gas law equation to solve for

#V_2 = P_1/P_2 * T_2/T_1 * V_1#

You know that

#P_2 = P_1/2 -># the pressure ishalved#T_2 = 2 * T_1 -># the temperature isdoubled

This means that you have

#V_2 = color(red)(cancel(color(black)(P_1)))/(1/2 * color(red)(cancel(color(black)(P_1)))) * (2 * color(red)(cancel(color(black)(T_1))))/color(red)(cancel(color(black)(T_1))) * V_1#

#V_2 = 2 * 2 * V_1 = 4 * V_1#

Indeed, both effects combine to cause an **increase** in volume.

The volume of the gas will thus be

#V_2 = 4 * "4.2 L" = "16.8 L"#

Rounded to two sig figs, the answer will be

#V_2 = color(green)("17 L")#