# How would the volume evolve if a 13000*L of gas under a 1.38*atm pressure were subjected to a 250*atm pressure?

Jul 21, 2017

The new volume would have to be smaller.........

#### Explanation:

We use old Boyle's law, which holds that $P \propto \frac{1}{V}$

And thus $P = \frac{k}{V}$, where $k$ is some constant.......and if we solve for $k$ under different conditions of pressure and volume (but constant temperature), then..........

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$....

And thus ${V}_{2} = \frac{{P}_{1} {V}_{1}}{P} _ 2 = \frac{1.38 \cdot a t m \times 13000 \cdot L}{250 \cdot a t m}$

$71.76 \cdot L$.

As regards the experimental setup, the experiment would have to be performed in a huge piston..... Perhaps the apparatus is not physically reasonable.