Question

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?

Answer 1

`"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`

Answer 2

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