A container with a volume of 16 L contains a gas with a temperature of 210^o K. If the temperature of the gas changes to 420 ^o K without any change in pressure, what must the container's new volume be?

Feb 25, 2016

The new volume is $32 L$

Explanation:

According to the Gas Equation we can write that $\frac{p \cdot V}{T} = c o n s t$

So now we know that the pressure $p$ does not change, so only temperaturw $\left(T\right)$ and volume$\left(V\right)$ changes

$\frac{p \cdot {V}_{1}}{T} _ 1 = \frac{p \cdot {V}_{2}}{T} _ 2$

if we divide by $p$ we have ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

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

${V}_{2} = \frac{16 \cdot 420}{210}$

${V}_{2} = 32 L$