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?

Question

2175 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-17T10:02:16

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 · 2018-03-17T11:28:00

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

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

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

$=24 \ "L"$

A container with a volume of 12 L12 L contains a gas with a temperature of 210 K210 K. If the temperature of the gas changes to 420 K420 K without any change in pressure, what must the container's new volume be?