If a gas in a closed container is pressurized from 15.0 atm to 16.0 atm and its original temperature was 25°C, what would be the final temperature of the gas in Kelvin?

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Answer 1 · 2018-01-06T18:48:21

317.867 K

Gay-Lussac's Law deals with pressure and temperature.

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

If we rearrange and get $T_{2}$ $T_2$ by itself, we get

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

Since temperature needs to be in Kelvin for all gas law calculations, $T_{1}$ $T_1$ = 298. That gives us the following

$T_{2} = \frac{16 \cdot 298}{15}$ $T_2= (16 * 298)/ 15$

That answer = 317.866666666666667