# If the initial pressure of a system was 1.00 atm and the volume was halved and the temperature in Kelvin was tripled, what is the ﬁnal pressure?

##### 1 Answer

#### Answer:

#### Explanation:

Even without doing any calculation, you can say that the pressure of the sample will **increase** by factor of

Here's why that is the case.

The idea here is to determine if the *change in volume* and the *change in temperature* will **counteract each other** or not.

The way to do that is to keep one *constant* and observe what happens to the pressure when the second changes, and vice versa.

So, let's say that temperature **is kept constant** and volume is halved. As you know, pressure and volume have an **inverse relationship** when temperature and number of moles of gas are kept constant - this is known as Boyle's Law.

#color(blue)(P * V = "constant")#

This means that a **reduction** in volume by a factor of **increase** in pressure by the same factor of

Now let's say that volume **is kep constant** and temperature is tripled. As you know, pressure and temperature have a **direct relationship** when volume and number of moles of gas are kept constant - this is known as Gay Lussac's Law.

#color(blue)(P/T = "constant")#

This means that an **increase** in temperature by a factor of **increase** in pressure by the same factor of

This tells you that *decreasing the volume* **and** *increasing the temperature* will lead to an overall **increase** in pressure by a factor of

You can prove this by using the combined gas law equation

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

Rearrange to solve for

#V_2 = V_1/2 -># the volume ishalved

#T_2 = 3 * T_1 -># the temperature istripled

You will get

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

#P_2 = color(red)(cancel(color(black)(V_1))) * 2/(color(red)(cancel(color(black)(V_1)))) * (3 * color(red)(cancel(color(black)(T_1))))/(color(red)(cancel(color(black)(T_1)))) * P_1#

#P_2 = 6 * P_1#

Therefore,

#P_2 = 6 * "1.00 atm" = color(green)("6.00 atm")#