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The pressure of a gas is reduced from 1200.0 mm hg to 850.0 mm hg as the volume of its container is increased by moving a piston from 85.0 ml to 350.0 ml. what would the final temperature be if the original temperature was 90.0 °c?

Respuesta :

jushmk
From gas laws:
[tex] \frac{PV}{T} = Constant[/tex]

Therefore,
[tex] \frac{ P_{1} V_{1} }{ T_{1} } = \frac{ P_{2} V_{2} }{ T_{2} } [/tex]

P1 = 1200 mm
V1 = 85 ml
T1 = 90°C = 363.15 K
P2 = 850 mm
V2 = 350 ml
T2 = ?

Substituting;
[tex] T_{2} = \frac{ P_{2} V_{2} T_{1} }{ P^{1} V^{1} } = \frac{850*350*363.15}{1200*85} = 1059.19 K[/tex] = 786.04 °C

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