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 #360 ^o K# without any change in pressure, what must the container's new volume be?

2 Answers
Oct 12, 2016

Answer:

#"The final volume of gas is "V_f=5.96L#

Explanation:

#V_i:"initial volume of gas ;"V_i=8L#

#T_i:"initial temperature of gas ;"T_i=210^o C=273+210=483^o K#

#T_f:"final temperature of gas ;"T_f=360^o K#

#V_i/T_i=V_f/T_f#

#8/483=V_f/360#

#8*360=483*V_f#

#V_f=(8*360)/483#

#V_f=5.96L#

Oct 12, 2016

Answer:

The new volume will be 6 L.

Explanation:

This is an example of Charles' law, which states that the volume of a gas held at constant pressure is directly proportional to its temperature in Kelvins.

The equation to use is #V_1/T_1=V_2/T_2#

Known Values
#V_1="8 L"#
#T_1="210"^@"C"+"273.15"="483 K"#
#T_2="360 K"#

Unknown
#V_2#

Solution
Rearrange the equation to isolate #V_2#. Substitute the known values into the equation and solve.

#V_2=(V_1T_2)/T_1#

#V_2=(8"L"xx360cancel"K")/(483cancel"K")="6 L"# rounded to one significant figure