# The gas inside of a container exerts 15 Pa of pressure and is at a temperature of 650 ^o K. If the temperature of the gas changes to 240 ^oC with no change in the container's volume, what is the new pressure of the gas?

Mar 9, 2017

$\approx 11.8 \text{Pa}$

#### Explanation:

Use the ideal gas law:

$p V = n R T$

Leaving out the constants/ things that don't change, this then boils down to:

$p \propto T$ or $\frac{p}{T} = c o n s t$, aka Gay-Lussac's Law

Remembering that $T$ is the absolute temperature in Kelvin, you then have:

${p}_{2} = {T}_{2} {p}_{1} / {T}_{1} = \left(240 + 273\right) \cdot \frac{15}{650} \approx 11.8 \text{Pa}$