# Fifty liters of gas is kept at a temperature of 200 K and under pressure of 15 atm. The temperature of the gas is increased to 400 K. The pressure is decreased to 7.5 atm. What is the resulting volume of the gas?

Feb 14, 2016

$\text{200 L}$

#### Explanation:

Even without doing any calculations, you should be able to look at the information given and predict that the volume of the gas will increase after temperature is increased and pressure is decreased.

In fact, these changes in pressure and temperature will "work together" to produce a more significant increase in volume.

So, you know that the sample starts at a pressure of $\text{15 atm}$ and a an absolute temperature of $\text{200 K}$. Under these conditions, the gas is said to occupy a volume of $\text{50 L}$.

Now, what would happen if you were to increase the temperature to $\text{400 K}$, but keep the pressure constant?

As you know, volume and temperature have a direct relationship when pressure and number of moles are kept constant $\to$ this is known as Charles' Law.

Mathematically, this is written as

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

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

Since you're essentially doubling the temperature of the gas, you have ${T}_{2} = 2 {T}_{1}$, which means that the volume changes to

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

Next, imagine that you're back to the original sample of gas. What would happen if you were to decrease the pressure to $\text{7.5 atm}$, but keep the temperature constant?.

When temperature and number of moles are kept constant, pressure and volume have an inverse relationship $\to$ this is known as Boyle's Law.

Mathematically, you can write this as

$\textcolor{b l u e}{{P}_{1} {V}_{1} = {P}_{2} {V}_{2}}$

In your case, the pressure is halved, so you have ${P}_{2} = \frac{1}{2} {P}_{1}$.

This will give you

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

Once again, the volume is doubled. This means that if you double the temperature and halve the pressure, the volume will increase by a factor of $2 \times 2 = 4$.

This is what the combined gas law equation is all about

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

Rearrange to solve for ${V}_{2}$

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

Plug in your values to confirm that the volume increases by a factor of $4$

V_2 = (15 color(red)(cancel(color(black)("atm"))))/(7.5color(red)(cancel(color(black)("atm")))) * (400color(red)(cancel(color(black)("K"))))/(200color(red)(cancel(color(black)("K")))) * "50 L" = color(green)("200 L")