A container with a volume of $6 L$ contains a gas with a temperature of $270^o K$ . If the temperature of the gas changes to $720 ^o K$ without any change in pressure, what must the container's new volume be?

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Answer 1 · 2017-01-30T08:23:45

The volume is $=16L$

We apply Charles' Law

$V_1/T_1=V_2/T_2$

$V_1=6L$

$T_1=270K$

$T_2=720K$

$V_2=T_2/T_1V_1$

$=720/270*6$

$=16L$

A container with a volume of 6 L6 L contains a gas with a temperature of 270^o K270^o K. If the temperature of the gas changes to 720 ^o K720 ^o K without any change in pressure, what must the container's new volume be?