Question

A constant temperature, a `12*L` volume of gas in a piston, is compressed to `3*L`. If the original pressure was `41*kPa`, what is the final pressure?

Answer 1

The pressure is `=164kPa`

Explanation:

We apply Boyle's Law

`P_1V_1=P_2V_2`, the temperature is constant

The initial pressure is `P_1=41kPa`

The initial volume is `V_1=12L`

The final volume is `V_2=3L`

The final pressure is

`P_2=V_1/V_2*P_1=12/3*41=164kPa`

Answer 2

Well, we use old Boyle's law.........and get `P_2=164*kPa`.

Explanation:

At constant temperature, and constant amount of gas, the product `PxxV=k`, where `k` is some constant....

And thus, if we solve for `k` at different conditions of volume and pressure, then `P_1V_1=P_2V_2`.

The utility of this equation is that we can use whatever whack units of volume and pressure we like, `"pints, pounds per square inch, torr, atmospheres, quarts"`, as long as we are consistent.

And so...`P_2=(P_1V_1)/V_2=(41*kPaxx12*L)/(3*L)=164*kPa`