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Keeping the number of moles constant, what is the final pressure in atm if the temperature was cooled from -91.5°C to -239°C, the volume decreased from 3.41L to 3.3L and the original pressure was 3990 torr?

Change celcius to Kelvin.

Respuesta :

Answer:

P2 = 0.935 atm

Explanation:

initial conditions:

∴ V1 = 3.41 L

∴ T1 = - 91.5 °C + 273 = 181.5 K

∴ P1 = 3990 torr * ( atm / 760 torr ) = 5.25 atm

  • PV = RTn

∴ R = 0.082 atm.L / K.mol

⇒ n = P1V1 / RT1 = ((5.25 atm)*(3.41 L)) / ((0.082 atm.L/K.mol)*(181.5 K))

⇒ n = 1.107 mol

final conditions:

∴ V2 = 3.3 L

∴ T2 = - 239 °C = 34 K

∴ n = 1.107 mol

⇒ P2 = nRT2 / V2

⇒ P2 = ((1.107 mol)*(0.082 atm.L/K.mol)*(34 K)) / 3.3 L

⇒ P2 = 0.935 atm

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