Question

The volume of a balloon is 6.52 L at 32.5 ºC and 762 mm Hg pressure. What volume in mL will the balloon have at standard temperature and pressure (STP)?  

Answer 1

Not too far from the initial volume....

Explanation:

The combined gas law holds that `(P_1V_1)/T_1=(P_2V_2)/T_2`

`V_2=(P_1V_1T_2)/(P_2T_1)`

We know that `760*mm*Hg-=1*atm`, `T=273.15*K` at `"STP"` (well, depending on your syllabus so check this....!)

`V_2=((760*mm*Hg)/(760*mm*Hg*atm^-1)xx6.52*Lxx273.15*K)/((762*mm*Hg)/(760*mm*Hg*atm^-1)xx305.7*K)=??*L`

And of course `1*L-=1000*mL`......

Reasonably the volume REDUCES slightly in that the temperature DROPS, and the pressure remains more or less constant.....