# 3.761xx10^1color(white)(.)"L" of an ideal gas is in a balloon at 1.000xx10^0 color(white)(.)"atm". The weather changes, and the volume of the balloon changes to 5.00xx10^1 "L". What is the new pressure, assuming no change in temperature?

Apr 26, 2017

The new pressure is $7.52 \times {10}^{- 1} \textcolor{w h i t e}{.} \text{atm}$.

#### Explanation:

This is an example of Boyle's law , which states that the volume of a gas held at constant amount and temperature, varies inversely with the pressure. This means that if the pressure increases, the volume decreases, and vice-versa. The equation for this law is:

"P_1V_1=P_2V_2

where $P$ is pressure and $V$is volume.

Organize the information:

Known
${P}_{1} = 1.000 \times {10}^{0} \text{atm}$
${V}_{1} = 3.761 \times {10}^{1} \text{L}$
${V}_{2} = 5.00 \times {10}^{1} \text{L}$

Unknown: ${P}_{2}$

Solution
Rearrange the equation to isolate ${P}_{2}$, substitute the known values into the equation and solve.

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

P_2=(1.000xx10^0"atm"xx3.761xx10^1color(red)cancel(color(black)"L"))/(5.00xx10^1color(red)cancel(color(black)"L"))=7.52xx10^(-1)"atm" rounded to three significant figures