A container with a volume of #15 L# contains a gas with a temperature of #290^o K#. If the temperature of the gas changes to #350 ^o K# without any change in pressure, what must the container's new volume be?

Question

A container with a volume of #15 L# contains a gas with a temperature of #290^o   K#.  If the temperature of the gas changes to #350     ^o    K# without any change in pressure, what must the container's new volume be?

Elena Albarran · Accepted Answer

#approx18.1L#.

According to ideal gas law, #P*V=n*R*T# Since, #P, n, R# are constant throughout the process, #V/T=(n*R)/P=constant# Therefore, #VpropT# #V_"initial"/T_"initial"=V_"final"/T_"final"# So, #V_"final"=(V_"initial"*T_"final")/T_"initial"=(15*350)/290approx18.1L# The answer is #18.1L#

Amelia Adams · Answer

#V_2=18.1 L# #P="constant"# #V_1=15L# #T_1=290 ^o K# #T_2=350 ^o K# #V_2=?# #V_1/T_1=V_2/T_2# #15/290=V_2/350# #V_2=(15*350)/290# #V_2=18.1 L#

Nathan Ashcroft · Answer

undefined Using the combined gas law, $ P_1V_1/T_1 = P_2V_2/T_2 $, where $ P $ is pressure, $ V $ is volume, and $ T $ is temperature:

Given:
$ V_1 = 15 \, \text{L} $
$ T_1 = 290 \, \text{K} $
$ T_2 = 350 \, \text{K} $

Since pressure remains constant, $ P_1 = P_2 $.

Rearranging the equation to solve for $ V_2 $:

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

$ V_2 = \frac{15 \times 350}{290} $

$ V_2 \approx 18.45 \, \text{L} $

Therefore, the new volume of the container should be approximately 18.45 L.