Question

A helium-filled balloon has a volume of 100 L at 25°C and 900 mmHg. What volume will it have at 180 kPa and 20°C?

Answer 1

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

Explanation:

The temperatures are quoted in `"degrees Kelvin"`. Of course, the twist introduced is the pressure value. We know that `1*atm` will support a column of mercury `760*mm` high, and the pressure measurement here is `900*mm`. This is not very realistic, as no one would measure those pressures with a mercury manometer. (Why not? Well. have you ever tried to clean up a mercury spill? Putting mercury under pressure is just asking for trouble.)

Thus `P_1=900*mm*Hg=(900*mm*Hg)/(760*mm*Hg*atm^1)` `=` `1.18*atm`

And `P_2=180*kPa=(180*kPa)/(101.32*kPa*atm^-1)` `=` `1.78*atm`

You now have consistent units, and I leave the calculation as an exercise.