A container with a volume of #19 L# contains a gas with a temperature of #220^o C#. If the temperature of the gas changes to #320 ^o K# without any change in pressure, what must the container's new volume be?

Question

1149 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-10T14:17:33

#12.33L = 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= 19L#
#T_1 = 220^oC# Convert to Kelvin #220 +273 = 493K#
#V_2=?#
#T_2=320K#

#V_1/T_1 =V_2/T_2#

#(19L)/(493K) =(V_2)/(320K)#

#(19L)/(493cancelK) * (320cancelK) =(V_2)/cancel(320K) * cancel(320K)#

#12.33L = V_2#