# If 7/5 L of a gas at room temperature exerts a pressure of 3 kPa on its container, what pressure will the gas exert if the container's volume changes to 5/3 L?

Feb 19, 2016

Start with the Combined Gas Law: $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

Temperature is constant, so $\left({P}_{1} {V}_{1}\right) = \left({P}_{2} {V}_{2}\right) \to {P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = \frac{3 \cdot \left(\frac{7}{5}\right)}{\frac{5}{3}} = \frac{\frac{21}{5}}{\frac{5}{3}} = \frac{36}{25}$ $k P a$

#### Explanation:

We could memorise Boyle's, Charles' and Gay-Lussac's Laws, which relate various combinations of pressure, volume and temperature, but I find it simpler to just memorise the Combined Gas Law and then eliminate the thing that is kept constant.

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

In this case, the temperature is kept constant, so we can ignore it:

$\left({P}_{1} {V}_{1}\right) = \left({P}_{2} {V}_{2}\right)$

Rearranging:

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