The gas inside of a container exerts #"24 Pa"# of pressure and is at a temperature of #"320 K"#. If the pressure in the container changes to #"32 Pa"# with no change in the container's volume, what is the new temperature of the gas?

1 Answer
Apr 2, 2016

Answer:

The final temperature is 430 K.

Explanation:

This is an example of Guy-Lussac's law, which states that as long as volume is constant, the pressure and temperature of a gas are directly proportional. This means that if the pressure goes up, so does the temperature and vice-versa. The equation for Gay-Lussac's law is #P_1/T_1=P_2/T_2#, where #P_1# is the initial temperature, #T_1# is the initial temperature, #P_2# is the final temperature, and #T_2# is the final temperature.

Given
#P_1="24 Pa"#
#T_1="320 K"#
#P_2="32 Pa"#

Unknown
#T_2#

Solution
Rearrange the equation to isolate #T_2#. Substitute the given values into the equation and solve.

#P_1/T_1=P_2/T_2#

#T_2=(P_2T_1)/P_1#

#T_2=(32cancel"Pa"xx320"K")/(24cancel"Pa")="430 K"# rounded to two significant figures