# 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?

Mar 17, 2018

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

So,we have,

$\frac{V}{T} = k$ where, $k$ is a constant

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

$k = \frac{12}{210}$

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

Then,we get,

$\frac{V '}{420} = k = \frac{12}{210}$

So,V'=(12/210)×420=24L

Mar 17, 2018

The new volume is $24$ liters or $24 \setminus \text{L}$.

#### Explanation:

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

$V \propto T$

or

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

Solving for ${V}_{2}$, we get:

${V}_{2} = {T}_{2} \cdot {V}_{1} / {T}_{1}$

Plugging in the given values, we find that

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

$= 24 \setminus \text{L}$