# A container with a volume of 18 L contains a gas with a temperature of 150^o K. If the temperature of the gas changes to 120 ^o K without any change in pressure, what must the container's new volume be?

Dec 22, 2017

$V = 14 \text{L}$

#### Explanation:

The ideal gas law gives:

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

Which for constant pressure becomes:

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

We can solve for ${V}_{2}$, the container's final volume:

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

We are given that:

• ${V}_{1} = 18 \text{L}$
• ${T}_{1} = 150 \text{K}$
• ${T}_{2} = 120 \text{K}$

Substituting these values into the equation above,

${V}_{2} = \frac{18 \cdot 120}{150}$

$= 14.4 \text{L}$

Or, ${V}_{2} \approx 14 \text{L}$