# 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"# ?

##### 1 Answer

#### Answer:

#### Explanation:

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

#color(blue)(ul(color(black)((P_1V_1)/T_1 = (P_2V_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

#color(blue)(ul(color(black)(T["K"] = t[""^@"C"] + 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_1V_1)/T_1 = (P_2V_2)/T_2 implies P_2 = V_1/V_2 * T_2/T_1 * 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**.