At a pressure 47 kPa, the gas in a cylinder has a volume of 12 liters. Assuming temperature remains the same, if the volume of the gas is decreased to 8 liters, what is the new pressure?

Aug 26, 2017

Boyles Law problem. Empirical relationship $P \propto \left(\frac{1}{V}\right)$ => For conditions 1 & 2 => ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$. Substitute given pressure & volume values and solve for the unknown. => 70.5 kPa.

Explanation:

Boyles Law is the only inverse relationship between gas phase variable of the Empirical Gas Law set. That is ...

For the Empirical Gas Laws the primary variables considered are Pressure (P), Volume (V). Temperature (T) and Mass (n). The macrolevel relationships are defined in terms of 4 'named expressions. Such are ...
Boyles Law => $P \propto \frac{1}{V}$ (mass & Temp are constant)
=> Decreasing V => Increases P (Inverse relationship)

Charles Law => $V \propto T$ (pressure & mass are constant)
=> Increasing T => Increasing V (Direct Relationship)

Gay-Lussac Law => $P \propto T$ (mass and Volume are constant)
=> Increasing T => Increasing P (Direct Relationship)

Avogadro's Law => $V \propto n$ (pressure and Temperature are constant) (Direct Relationship)

Boyles Law therefore applies to the posted problem and for two conditions given, the following expression is applied ...

${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$
Substituting given data ...
${P}_{1} = 47 k P a$
${V}_{1} = 12 L i t e r s$
P_2 = ? "(unknown)"
${V}_{2} = 8 L i t e r s$

Substituting and solving for ${P}_{2}$ ...

=> $\left(47 k P a\right) \left(12 L i t e r s\right) = \left({P}_{2}\right) \left(8 L i t e r s\right)$

=> ${P}_{2} = \left(47 k P a\right) \frac{12 \cancel{L}}{8 \cancel{L}}$ = $70.5 k P a$
(Note: The decreased Volume (12L => 8L) resulted in an increase in pressure (47kPa => 70.5kPa); consistent with Boyles Law relationship.)