更新1:
A sample contains .632 mol of O2 and .758 mol of H2O in a 5.53 L container.? Edit
A sample contains .632 mol of O2 and .758 mol of H2O in a 5.56 L container.?
2016-05-31 12:13 am
Calculate the partial press. of each gas then calculate the total press. at 78 C
回答 (1)
2016-05-31 12:34 am
I think it should be H₂ instead of H₂O. Under such conditions, H₂O exists as a liquid, but not a gas.
PV = nRT
Then, P = nRT/V
Partial pressure of O₂, P(O₂)
= 0.632 × 0.0821 × (273 + 78) / 5.53 atm
= 3.29 atm
Partial pressure of H₂, P(H₂)
= 0.758 × 0.0821 × (273 + 78) / 5.53 atm
= 3.95 atm
Total pressure
= (3.29 + 3.95) atm
= 7.24 atm
OR: Total pressure
= (0.632 + 0.758) × 0.0821 × (273 + 78) / 5.53 atm
= 7.24 atm
====
If it is really H₂O, but not H₂ :
Partial pressure of H₂O, P(H₂O)
= Vapor pressure of water at 78°C
= 327.3 torr
= 327.3/760 atm
= 0.431 atm
Total pressure
= (3.29 + 0.431) atm
= 3.72 atm
PV = nRT
Then, P = nRT/V
Partial pressure of O₂, P(O₂)
= 0.632 × 0.0821 × (273 + 78) / 5.53 atm
= 3.29 atm
Partial pressure of H₂, P(H₂)
= 0.758 × 0.0821 × (273 + 78) / 5.53 atm
= 3.95 atm
Total pressure
= (3.29 + 3.95) atm
= 7.24 atm
OR: Total pressure
= (0.632 + 0.758) × 0.0821 × (273 + 78) / 5.53 atm
= 7.24 atm
====
If it is really H₂O, but not H₂ :
Partial pressure of H₂O, P(H₂O)
= Vapor pressure of water at 78°C
= 327.3 torr
= 327.3/760 atm
= 0.431 atm
Total pressure
= (3.29 + 0.431) atm
= 3.72 atm
收錄日期: 2021-04-18 15:03:19
原文連結 [永久失效]:
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