The gas inside of a container exerts #9 Pa# of pressure and is at a temperature of #35 ^o K#. If the temperature of the gas changes to #20 ^oK# with no change in the container's volume, what is the new pressure of the gas?

Question

The gas inside of a container exerts #9   Pa# of pressure and is at a temperature of #35 ^o K#.  If the temperature of the gas changes to #20 ^oK# with no change in the container's volume, what is the new pressure of the gas?

Adrian Ayala · Accepted Answer

#5# #"Pa"#

To solve this problem, we can use the pressure-temperature relationship of gases, illustrated by Gay-Lussac's law:

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

Let's rearrange this equation to solve for the final pressure, #P_2#:

#P_2 = (P_1T_2)/(T_1)#

Plugging in known values, we have

#P_2 = ((9"Pa")(20cancel("K")))/(35cancel("K")) = color(red)(5)# #color(red)("Pa"#

rounded to one significant figure, the lowest amount given in the problem.

Adrian Ayala · Answer

undefined You can use the ideal gas law to solve this problem: $ P_1/T_1 = P_2/T_2 $, where $ P_1 $ and $ T_1 $ are the initial pressure and temperature, and $ P_2 $ and $ T_2 $ are the final pressure and temperature.

Given $ P_1 = 9 \, \text{Pa} $ and $ T_1 = 35 \, \text{K} $, and $ T_2 = 20 \, \text{K} $ (with no change in volume), you can rearrange the formula to solve for $ P_2 $.

$ P_2 = P_1 \times \frac{T_2}{T_1} $

$ P_2 = 9 \, \text{Pa} \times \frac{20 \, \text{K}}{35 \, \text{K}} $

$ P_2 \approx 5.14 \, \text{Pa} $