The gas inside of a container exerts #27 Pa# of pressure and is at a temperature of #150 ^o K#. If the pressure in the container changes to #30 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 #27   Pa# of pressure and is at a temperature of #150 ^o K#.  If the pressure in the container changes to #30 Pa# with no change in the container's volume, what is the new temperature of the gas?

Isabella Alvarez · Accepted Answer

#T_2 = 170# #"K"#

We'r asked to find the new temperature of a gas,after it is subjected to a pressure change with no change in volume.

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

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

Here,

#P_1 = 27# #"Pa"#

#T_1 = 150# #"K"#

#P_2 = 30# #"Pa"#

#T_2 = ?#

Plugging in known values and solving for #T_2#, we have

#T_2 = (P_2T_1)/(P_1) = ((30cancel("Pa"))(150color(white)(l)"K"))/(27cancel("Pa")) = color(red)(170# #color(red)("K"#

rounded to #2# significant figures.

Isabella Alvarez · Answer

undefined To find the new temperature of the gas when the pressure changes from 27 Pa to 30 Pa with no change in volume, we can use the ideal gas law:

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

Given:
$$ P_1 = 27 \, \text{Pa} $$
$$ T_1 = 150 \, \text{K} $$
$$ P_2 = 30 \, \text{Pa} $$
$$ V_1 = V_2 \, (\text{no change in volume}) $$

Substitute the values into the formula and solve for $ T_2 $.