A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be?

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A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be?

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Answer 1 · 2017-03-18T04:12:20

The final temperature will be $\approx 200 . K$ $~~200color(white)(.)"K"$ .

Explanation:

This question involves Gay-Lussac's law to answer, which states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

This means that as the pressure goes up, the temperature also goes up, and vice-versa.

The equation to use is:

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

where $P$ $P$ is pressure in atm, and $T$ $T$ is temperature in Kelvins.

Given/Known Information
$P_{1} = 3.8 atm$ $P_1="3.8 atm"$
$T_{2} = 500 K$ $T_2="500 K"$
$P_{2} = 1.2 atm$ $P_2="1.2 atm"$

Unknown: $T_{2}$ $T_2$

Solution
Rearrange the equation to isolate $T_{2}$ $T_2$ .

$T_{2} = \frac{P_{2} T_{1}}{P_{1}}$ $T_2=(P_2T_1)/(P_1)$

$T_2=(1.2color(red)cancel(color(black)("atm"))xx500 "K")/(3.8color(red)cancel(color(black)("atm")))~~"200 K"$ rounded to 1 significant figure

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A gas is held at 3.8 atm and 500 K. If the pressure is then decreased to 1.2 atm, what will the new temperature be?

1 Answer

Explanation:

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