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Alinara [238K]
3 years ago
15

if a gas is held in 3.60 L, 298 K AND 800. mmHg, what is the new pressure of the gas if the volume us decreased by half and the

tempature is decreased to 250. K
Chemistry
2 answers:
sleet_krkn [62]3 years ago
6 0

Answer:

Explanation:

By Ideal Gas Law, P1*V1 / T1 = P2*V2 / T2

So new pressure = (P1*V1 / T1) / (V2 / T2)

= P1*V1*T2 / T1*V2

= 800*3.6*298 / 250*1.8

= 1907.2 mmHg

AlexFokin [52]3 years ago
6 0

Apply combine gas law

\\ \rm\Rrightarrow \dfrac{P1V1}{T1}=\dfrac{P2V2}{T2}

\\ \rm\Rrightarrow P2=\dfrac{P1V1T2}{V2T1}

\\ \rm\Rrightarrow P2=\dfrac{800(3.6)(250)}{1.8(298)}

\\ \rm\Rrightarrow P2=720000/536.4

\\ \rm\Rrightarrow P2=1341.7 mmHg

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