The gas inside of a container exerts #18 Pa# of pressure and is at a temperature of #450 ^o K#. If the pressure in the container changes to #25 Pa# with no change in the container's volume, what is the new temperature of the gas?

1 Answer
Dec 25, 2017

Answer:

The new temperature will be #~~##"630 K"#.

Explanation:

This is an example of Guy-Lussac's law which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. So if the pressure increases, so will the temperature and vice versa.

The equation is:

#P_1/T_1=P_2/T_2#,

where:

#P# is pressure and #T# is temperature.

Organize the data.

Known

#P_1="18 Pa"#

#T_1 ="450 K"#

#P_2="25 Pa"#

Unknown

#T_2#

Solution

Rearrange the equation to isolate #T_2#. Plug in the known values and solve.

#T_2=(P_2T_1)/(P_1)#

#T_2=(25color(red)cancel(color(black)("Pa"))xx450"K")/(18color(red)cancel(color(black)("Pa")))="630 K"# rounded to two significant figures due to 18 Pa and 25 Pa