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Answer 1 · 2017-05-13T23:01:31

$- 129^{o} C$ $-129 ^oC$

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

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ $(P_1)/(T_1) = (P_2)/(T_2)$

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

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

We can rearrange this equation to solve for the final temperature, $T_{2}$ $T_2$ :

$T_{2} = (T_{1}) \frac{P_{2}}{P_{1}}$ $T_2 = (T_1)(P_2)/P_1$

Plug in known variables:

$T_{2} = 288 K \frac{225 m m H g}{450 m m H g} = 144 K$ $T_2 = 288 K(225 mm Hg)/(450 mm Hg) = 144 K$

Finally, we convert the Kelvin temperature back to Celsius:

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