# If 10/3 L of a gas at room temperature exerts a pressure of 42 kPa on its container, what pressure will the gas exert if the container's volume changes to 12/7 L?

Mar 21, 2018

The new pressure is $= 81.7 k P a$

#### Explanation:

Apply Boyle's Law

$\text{Pressure" xx"Volume"="Constant}$ at constant temperature

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

The initial pressure is ${P}_{1} = 42 k P a$

The initial volume is ${V}_{1} = \frac{10}{3} L$

The final volume is ${V}_{2} = \frac{12}{7} L$

The final pressure is

${P}_{2} = {V}_{1} / {V}_{2} \cdot {P}_{1} = \frac{\frac{10}{3}}{\frac{12}{7}} \cdot 42 = 81.7 k P a$

Mar 22, 2018

$81.7 \setminus \text{kPa}$

#### Explanation:

We use Boyle's law, if the volume and the number of moles is constant, and we get,

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

Now, we plug in the given values, and we get

$42 \setminus \text{kPa"*10/3color(red)cancelcolor(black)"L"=P_2*12/7color(red)cancelcolor(black)"L}$

${P}_{2} \approx 81.7 \setminus \text{kPa}$

So, the new pressure is approximately $81.7$ kilopascals.