# What happens to the volume of an ideal gas if the pressure halves and temperature doubles?

##### 1 Answer

Well... quadrupled.

Consider the ideal gas law...

#bb(PV = nRT)# where

#P# ,#V# ,#n# ,#R# , and#T# are pressure in#"atm"# , volume in#"L"# , mols in#"mol"# s, the universal gas constant in#"L"cdot"atm/mol"cdot"K"# , and temperature in#"K"# , respectively.

You should know how to manipulate this equation in general for an exam---this is fair play.

**METHOD 1**

*Volume is a state function, and hence one can break down two changes in related variables into two separate steps.*

Since the pressure halved and the temperature doubled, consider their stacked effects in two steps...

**Step 1:**halving the pressure at*constant temperature*means half the volume compression, so the volume**doubles**after step 1.**Step 2:**doubling the temperature at*constant pressure***doubles**the volume after step 2.

So, the volume approximately **quadruples**.

**METHOD 2**

Let

#1/2P (cV) = nR(2T)#

#color(blue)(barul(stackrel(" ")(|" "cPV = 4nRT" ")|))#

Thus, since **the volume quadrupled**.

**METHOD 3**

Another way to do this is to set initial and final states.

#P_1V_1 = nRT_1#

#P_2V_2 = nRT_2#

And we say, what is

#1/2P_1V_2 = 2nRT_1#

#=> color(blue)(barul(stackrel(" ")(|" "V_2 = (color(red)(4)nRT_1)/P_1" ")|))#

For

#=> color(blue)(barul(stackrel(" ")(|" "V_1 = (nRT_1)/P_1" ")|))#

By inspection, we see that **the volume has quadrupled**.