# Question #a4a6a

##### 1 Answer

#### Answer:

Its density doubles.

#### Explanation:

Before doing any calculation, try to predict what you *expect* to happen to the density of a gas if pressure is **increased** while temperature is **kept constant**.

As you know, density is defined as the *mass* per unit of volume. Since the mass of the gas is also kept **constant**, the only way to change its density is to change the volume it occupies.

Now, what happens to the volume of an ideal gas when pressure is **increased** at constant temperature and number of moles?

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

What that tells you is that *increasing* the pressure of the gas will cause its volume to **decrease**. Likewise, *decreasing* the pressure of the gas will cause its volume to **increase**.

So right from the start you can say that since the pressure of the gas is *increased*, the volume will *decrease*, which in turn will cause the density of the gas to **increase**, since now you have the *same mass of gas* in a **smaller volume**.

If you start with *initial state*, you can use Boyle's Law to write

#color(blue)(P_1 * V_1 = P_2 * V_2) -># the equation tha describesBoyle's Law

Here *final state*.

In your case, the pressure is **doubled**, so you can say that

#P_2 = 2 * P_1#

Plug this into the equation and solve for

#V_2 = P_1/P_2 * V_1#

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

The density of the gas at the initial state was

#rho_1 = m/V_1" "# , where

The density of the gas at the final state will be

#rho_2 = m/color(purple)(V_2) = m * 1/(color(purple)(1/2 * V_1)) = 2 * overbrace(m/V_1)^(color(red)(=rho_1))#

Therefore,

#rho_2 = color(green)(2 * rho_1)#

So, you can now say that **doubling** the pressure of the gas will **halve** its volume, which in turn will **double** its density.