# 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?

Aug 11, 2017

The pressure is $= 164 k P a$

#### Explanation:

We apply Boyle's Law

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$, the temperature is constant

The initial pressure is ${P}_{1} = 41 k P a$

The initial volume is ${V}_{1} = 12 L$

The final volume is ${V}_{2} = 3 L$

The final pressure is

${P}_{2} = {V}_{1} / {V}_{2} \cdot {P}_{1} = \frac{12}{3} \cdot 41 = 164 k P a$

Aug 11, 2017

Well, we use old Boyle's law.........and get ${P}_{2} = 164 \cdot k P a$.

#### Explanation:

At constant temperature, and constant amount of gas, the product $P \times V = k$, where $k$ is some constant....

And thus, if we solve for $k$ at different conditions of volume and pressure, then ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$.

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

And so...${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = \frac{41 \cdot k P a \times 12 \cdot L}{3 \cdot L} = 164 \cdot k P a$