A container with a volume of #4 L# contains a gas with a temperature of #770^o K#. If the temperature of the gas changes to #420^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #4 L# contains a gas with a temperature of #770^o   K#.  If the temperature of the gas changes to #420^o    K# without any change in pressure, what must the container's new volume be?

Leo Abeyta · Accepted Answer

#V_2=2,18L# #"given data:"# #V_1=4L# #T_1=770^o K# #T_2=420^o K# #V_2=?# #"solution:"# #V_1/T_1=V_2/T_2# #4/770=V_2/420# #V_2=(4*420)/770# #V_2=2,18L#

Nora Arzate · Answer

undefined Using the combined gas law formula $ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} $, where $ P $ is pressure, $ V $ is volume, and $ T $ is temperature, we can solve for the new volume:

$ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} $

Given:
$ V_1 = 4 \, \text{L} $
$ T_1 = 770 \, \text{K} $
$ T_2 = 420 \, \text{K} $
$ P_1 = P_2 $ (pressure remains constant)

$ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} $

$ \frac{P_1 \times 4}{770} = \frac{P_2 \times V_2}{420} $

$ \frac{4P_1}{770} = \frac{P_2 \times V_2}{420} $

$ \frac{4P_1 \times 420}{770} = P_2 \times V_2 $

$ V_2 = \frac{4P_1 \times 420}{770} $

Since $ P_1 = P_2 $, we can replace $ P_1 $ with $ P_2 $:

$ V_2 = \frac{4P_2 \times 420}{770} $

$ V_2 = \frac{1680}{770} $

$ V_2 ≈ 2.18 \, \text{L} $

Therefore, the container's new volume should be approximately $ 2.18 \, \text{L} $.