# How do you calculate the final pressure of a gas that is compressed from a volume of "20.0 dm"^3 to "10.0 dm"^3 and cooled from 100^@"C" to 25^@"C" if the initial pressure is "1 bar" ?

Mar 1, 2017

$\text{1.6 bar}$

#### Explanation:

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

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

Here

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

Before doing anything else, make sure that you convert the temperatures from degrees Celsius to Kelvin by using the fact that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{T \left[\text{K"] = t[""^@"C}\right] + 273.15}}}$

Now, the idea here is that decreasing the volume of the gas will cause the pressure to increase. On the other hand, decreasing the temperature of the gas will cause its pressure to decrease.

You can thus say that the change in volume and the change in temperature will "compete" each other, i.e. whichever change is more significant will determine if the pressure increases or decreases.

So, rearrange the combined gas law to solve for ${P}_{2}$

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

Plug in your values to find

P_2 = (20.0 color(red)(cancel(color(black)("dm"^3))))/(10.0color(red)(cancel(color(black)("dm"^3)))) * ((25 + 273.15)color(red)(cancel(color(black)(""^@"C"))))/((100 + 273.15)color(red)(cancel(color(black)(""^@"C")))) * "1 bar"

P_2 = 2 * 0.799 * "1 bar" = color(darkgreen)(ul(color(black)("1.6 bar")))

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the initial pressure of the gas.

As you can see, the decrease in volume was more significant than the decrease in temperature; as a result, the pressure of the gas increased.