Question

`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?

Answer 1

The new pressure is `7.52xx10^(-1)color(white)(.)"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=1.000xx10^0"atm"`
`V_1=3.761xx10^1 "L"`
`V_2=5.00xx10^1"L"`

Unknown: `P_2`

Solution
Rearrange the equation to isolate `P_2`, substitute the known values into the equation and solve.

`P_2=(P_1V_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