Question #12b3b

Like I said above, the same amount of the gas is squeezed into a smaller area with an increase in pressure.

Remember that pressure is measured by the amount of molecules of the gas hitting the sides of the wall in the container.

Because of this, when the pressure increases, so does the number of molecules hitting the wall of the container, and the best explanation for that is that the walls have come closer in, thus a smaller volume.

This can be proven using Boyle's Law, which is a form of the Combined gas law. The combined gas law states that: #(V_s P_s) / T_s = (V_f P_f) / T_f#
Where #V# stands for Volume, #P# stands for pressure, and #T# stands for Temperature. And the subscripts "s" and "f" stand for start and finish respectively. Since you did not mention anything about the temperature, it is safe to assume you are talking about that when the temperature is constant.

So the #T#s all cancel each other out, leaving Boyle's Law, which says #V_s P_s = V_f P_f#

Since you original question asks why when the pressure increases, the volume decreases, so let me demonstrate that for you.

Say we had #2 "Liters of gas"=Volume# at #1 atmosphere=Pressure#. These are both our starting measurements, so let's put them in for #V_s# and #P_s#

So now we have #2 L * 1 atm = V_f P_f#

And your situation says if we increase the pressure, the volume will decrease.

So let's change the starting pressure from 1 atm to 2 atm in the finishing pressure. So now we have #2 L * 1 atm = V_f * 2atm#

To solve for #V_f# we divide both sides by the 2 atm and then we are left with #1 Liter = V_f#

So if we increase the pressure of our example to 2 atm, our volume will decrease to 1 Liter.

So there is your explanation of why when you increase the pressure, the volume will decrease.

Hope it helps!