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?

Question

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?

Amelia Adams · Accepted Answer

#540"K"#

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"#

Isabella Alvarez · Answer

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

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.

Elena Albarran · Answer

undefined To find the new temperature of the gas, we can use the ideal gas law equation:

$$ P_1 \times V_1 / T_1 = P_2 \times V_2 / T_2 $$

Given:
$$ P_1 = 18 \, Pa $$
$$ T_1 = 360 \, K $$
$$ P_2 = 27 \, Pa $$
$$ V_1 = V_2 $$ (since the volume remains constant)

Solve for $ T_2 $:
$$ T_2 = (P_2 \times T_1) / P_1 $$

$$ T_2 = (27 \times 360) / 18 $$
$$ T_2 = 540 \, K $$

So, the new temperature of the gas is $ 540 \, K $.