# 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?

Jan 1, 2016

${P}_{2} = \text{6.00 atm}$

#### Explanation:

Even without doing any calculation, you can say that the pressure of the sample will increase by factor of $6$.

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.

$\textcolor{b l u e}{P \cdot V = \text{constant}}$

This means that a reduction in volume by a factor of $2$ will trigger an increase in pressure by the same factor of $2$.

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.

$\textcolor{b l u e}{\frac{P}{T} = \text{constant}}$

This means that an increase in temperature by a factor of $3$ will trigger an increase in pressure by the same factor of $3$.

This tells you that decreasing the volume and increasing the temperature will lead to an overall increase in pressure by a factor of $2 \times 3 = 6$.

You can prove this by using the combined gas law equation

$\textcolor{b l u e}{\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2} \text{ }$, where

${P}_{1}$, ${V}_{1}$, ${T}_{1}$ - the pressure, volume, and temperature of the gas at an initial state
${P}_{2}$, ${V}_{2}$, ${T}_{2}$ - the pressure, volume, and temperature of the gas at a final state

Rearrange to solve for ${P}_{2}$ and use

${V}_{2} = {V}_{1} / 2 \to$ the volume is halved

${T}_{2} = 3 \cdot {T}_{1} \to$ the temperature is tripled

You will get

${P}_{2} = {V}_{1} / {V}_{2} \cdot {T}_{2} / {T}_{1} \cdot {P}_{1}$

${P}_{2} = \textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}} \cdot \frac{2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{V}_{1}}}}} \cdot \frac{3 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{T}_{1}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{T}_{1}}}}} \cdot {P}_{1}$

${P}_{2} = 6 \cdot {P}_{1}$

Therefore,

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