Question

A container with a volume of `12 L` contains a gas with a temperature of `210 K`. If the temperature of the gas changes to `420 K` without any change in pressure, what must the container's new volume be?

Answer 1

Just apply Charle's law for constant pressure and mas of an ideal gas,

So,we have,

`V/T=k` where, `k` is a constant

So,we putting the initial values of `V` and `T` we get,

`k=12/210`

Now,if new volume is `V'` due to temperature `420K`

Then,we get,

`(V')/420=k=12/210`

So,`V'=(12/210)×420=24L`

Answer 2

The new volume is `24` liters or `24 \ "L"`.

Explanation:

Since there is no change in temperature and the number of moles, we use Charles's law, which states that

`VpropT`

or

`V_1/T_1=V_2/T_2`

Solving for `V_2`, we get:

`V_2=T_2*V_1/T_1`

Plugging in the given values, we find that

`V_2=420color(red)cancelcolor(black)"K"*(12 \ "L")/(210color(red)cancelcolor(black)"K")`

`=24 \ "L"`