The gas inside of a container exerts #"84 Pa"# of pressure and is at a temperature of #"320 K"#. If the pressure in the container changes to #"64 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 #"84 Pa"# of pressure and is at a temperature of #"320 K"#.  If the pressure in the container changes to #"64 Pa"# with no change in the container's volume, what is the new temperature of the gas?

Adrian Ayala · Accepted Answer

Using the ideal gas law to derive an equation is one method to accomplish this. #PV = nRT# where Then, we would say that if state #1# represented an initial state and state #2# a final state, we have #P_1 -> P_2# and #T_1 -> T_2#, but #V_1 = V_2 = V# and #n_1 = n_2 = n#. We can derive two ideal gas law relationships using this information: #P_1V = nRT_1#
#P_2V = nRT_2# Thus, we can divide these to obtain the new temperature: #P_1/P_2 = T_1/T_2# #\mathbf(T_2 = T_1*P_2/P_1)# So, now that we have the final equation, we can acquire #T_2# for an ideal gas. #color(blue)(T_2) = "320 K" xx (84/64)# #=# #color(blue)("420 K")#

Lily Adams · Answer

undefined To find the new temperature of the gas, use Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature when the volume and the amount of gas are constant.

Given:
Initial pressure, $ P_1 = 84 \, \text{Pa} $
Initial temperature, $ T_1 = 320 \, \text{K} $
Final pressure, $ P_2 = 64 \, \text{Pa} $

Using Gay-Lussac's Law:
$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $

Substituting the given values:
$ \frac{84}{320} = \frac{64}{T_2} $

Solving for $ T_2 $:
$ T_2 = \frac{64 \times 320}{84} $
$ T_2 = \frac{20480}{84} $
$ T_2 \approx 244.76 \, \text{K} $

Therefore, the new temperature of the gas is approximately 244.76 K.