A container with a volume of #8 L# contains a gas with a temperature of #210^o C#. If the temperature of the gas changes to #310 ^o K# without any change in pressure, what must the container's new volume be?

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Answer 1 · 2016-07-10T14:34:47

#5.13L = V_2#

To compare the volume and temperature change when pressure is held constant use Charles' Law #V_1/T_1 =V_2/T_2#

#V_1= 8L#
#T_1 = 210^oC# Convert to Kelvin #210 +273 = 483K#
#V_2=?#
#T_2=310K#

#V_1/T_1 =V_2/T_2#

#(8L)/(483K) =(V_2)/(310K)#

#(8L)/(483cancelK) * (310cancelK) =(V_2)/cancel(310K) * cancel(310K)#

#5.13L = V_2#