#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?

1 Answer
Apr 26, 2017

Answer:

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