The gas inside of a container exerts #18 Pa# of pressure and is at a temperature of #360 ^o K#. If the pressure in the container changes to #27 Pa# with no change in the container's volume, what is the new temperature of the gas?

2 Answers
May 23, 2017

Answer:

The final temperature, #T_2#, will be #"540 K"#.

Explanation:

This question is an example of Gay-Lussac's gas law, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. This means that if the pressure increases, so does the temperature, and vice-versa. The equation to use in order to answer your question is:

#P_1/T_1=P_2/T_2#

Organize your data.

Given

#P_1="18 Pa"#
#T_1="360 K"# #lArr# The degree symbol #(""^@")# is not used with Kelvins.
#P_2="27 Pa"#

Unknown: #T_2#

Solution
Rearrange the equation to isolate #T_2#. Insert your data into the equation and solve.

#T_2=(P_2T_1)/(P_1)#

#T_2=(27color(red)cancel(color(black)("Pa"))xx360"K")/(18color(red)cancel(color(black)("Pa")))="540 K"#

As you can see, as the pressure increased, so did the temperature.

May 23, 2017

Answer:

#540"K"#

Explanation:

We can use the temperature-pressure relationship of gases illustrated by Gay-Lussac's law:

#(P_1)/(T_1) = (P_2)/(T_2)#

At constant volume, the temperature and pressure of a fixed quantity of gas are directly proportional to each other, which is explained by the kinetic-molecular theory.

We can rearrange this equation to solve for the final temperature #T_2#:

#T_2 = (P_2T_1)/(P_1)#

and finally, plug in the known variables to find the temperature.

#T_2 = ((27"Pa")(360K))/(18"Pa") = color(red)(540"K"#