# If 9 L of a gas at room temperature exerts a pressure of 15 kPa on its container, what pressure will the gas exert if the container's volume changes to 4 L?

Mar 6, 2016

$33.75 k P a$

#### Explanation:

From the information given for this question, we can see that this kind of situation is involving Boyle's Law.

Boyle's Law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume .

In which from the definition, the equation is derived as;

$P \quad \propto \quad \frac{1}{V}$ or $P = \frac{k}{V}$ or $P V = k$

$P =$ Pressure of gas
$V =$ Volume of gas
$k =$ Constant

When there are two situations , given initial and final value of both pressure and volume, the equation is derived as;

${P}_{i} {V}_{i} = {P}_{f} {V}_{f}$

From the information given in this question;

${P}_{i} =$ Initial pressure of gas $= 15 k P a$
${P}_{f} =$ Final pressure of gas =?kPa
${V}_{i} =$ Initial volume of gas $= 9 L$
${V}_{f} =$ Final volume of gas $= 4 L$

Calculating ${P}_{f}$;

$\left(15 k P a\right) \left(9 L\right) = {P}_{f} \left(4 L\right)$

${P}_{f} = \frac{135 k P a \cancel{L}}{4 \cancel{L}}$

${P}_{f} = 33.75 k P a$