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?

Question

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?

Julia Aguilar · Accepted Answer

The final temperature is 430 K.

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

Nora Arzate · Answer

undefined You can use the ideal gas law, $ PV = nRT $, where $ P $ is pressure, $ V $ is volume, $ n $ is the number of moles of gas, $ R $ is the gas constant, and $ T $ is temperature. Since the volume remains constant, we can simplify the equation to $ 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 that $ P_1 = 24 \, \text{Pa} $, $ T_1 = 320 \, \text{K} $, and $ P_2 = 32 \, \text{Pa} $, we can rearrange the equation to solve for $ T_2 $:

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

$$ T_2 = \frac{{32 \, \text{Pa} \times 320 \, \text{K}}}{{24 \, \text{Pa}}} $$

$$ T_2 = 426.67 \, \text{K} $$

So, the new temperature of the gas is $ 426.67 \, \text{K} $.