# A sample of compressed methane has a volume of 648 mL at a pressure of 503 kPa. To what pressure would the methane have to be compressed in order to have a volume of 216 mL?

Mar 14, 2017

#### Answer:

The pressure is $= 1509 k P a$

#### Explanation:

We apply Boyle's Law

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

${P}_{1} = 503 k P a$

${V}_{1} = 648 m L$

${V}_{2} = 216 m L$

${P}_{2} = {V}_{1} / {V}_{2} \cdot {P}_{1}$

$= \frac{648}{216} \cdot 503 = 1509 k P a$

Mar 14, 2017

#### Answer:

Well, Boyle's law holds that ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$............

#### Explanation:

So we solve for P_2=(P_1V_1)/V_2=(503*kPaxx648*mL)/(216*mL)=??*kPa.

Clearly, the pressure increases, here almost threefold.