# Question #1b6aa

May 13, 2017

$- {129}^{o} C$

#### Explanation:

This question involves the pressure-temperature relationship of gases, and is illustrated by Gay-Lussac's law:

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

Remember that the temperature is the absolute temperature, so you have to convert the Celsius temperature to Kelvin:

${15}^{o} C + 273 = 288 K$

We can rearrange this equation to solve for the final temperature, ${T}_{2}$:

${T}_{2} = \left({T}_{1}\right) \frac{{P}_{2}}{P} _ 1$

Plug in known variables:

${T}_{2} = 288 K \frac{225 m m H g}{450 m m H g} = 144 K$

Finally, we convert the Kelvin temperature back to Celsius:

$144 K - 273 = - {129}^{o} C$