# 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?

Jun 30, 2016

The combined gas law holds that $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

#### Explanation:

The temperatures are quoted in $\text{degrees Kelvin}$. Of course, the twist introduced is the pressure value. We know that $1 \cdot a t m$ will support a column of mercury $760 \cdot m m$ high, and the pressure measurement here is $900 \cdot m m$. 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 ${P}_{1} = 900 \cdot m m \cdot H g = \frac{900 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{1}}$ $=$ $1.18 \cdot a t m$

And ${P}_{2} = 180 \cdot k P a = \frac{180 \cdot k P a}{101.32 \cdot k P a \cdot a t {m}^{-} 1}$ $=$ $1.78 \cdot a t m$

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