The gas inside of a container exerts #14 Pa# of pressure and is at a temperature of #220 ^o C#. If the temperature of the gas changes to #150 ^oC# with no change in the container's volume, what is the new pressure of the gas?

Question

The gas inside of a container exerts #14   Pa# of pressure and is at a temperature of #220 ^o C#.  If the temperature of the gas changes to #150 ^oC# with no change in the container's volume, what is the new pressure of the gas?

Layla Ahmed · Accepted Answer

The new pressure is #=3.4Pa#

We employ the Law of Gay Lussac.

#P_1/T_1=P_2/T_2#

The initial pressure is #P_1=14Pa#

The initial temperature is #T_1=220+273=493K#

The final temperature is #T_2=150+273=423K#

The last bit of pressure is

#P_2=T_2/T_1*P_1#

#=423/493*14#

#=3.4Pa#

Isabella Alvarez · Answer

undefined To find the new pressure of the gas, you can use the combined gas law, which states that for a given amount of gas at constant volume:

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

Where:
$ P_1 $ = initial pressure,
$ T_1 $ = initial temperature,
$ P_2 $ = final pressure,
$ T_2 $ = final temperature.

Rearranging the equation to solve for $ P_2 $, we get:

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

Substitute the given values:

$ P_2 = 14 \times \frac{150 + 273}{220 + 273} $

$ P_2 = 14 \times \frac{423}{493} $

$ P_2 \approx 12.02 \, \text{Pa} $

So, the new pressure of the gas is approximately 12.02 Pa.