ball that contained 25.0 L of air at 22C and 6.25 atm pressure was placed in an oven at a temperature of 100C. what is the new pressure?
回答 (2)
2018-04-09 10:56 am
Initial: P₁ = 6.25 atm, T₁ = (273 + 22) K = 295 K
New: P₂ = ? atm, T₂ = (273 + 100) K = 373 K
For a fixed amount of gas at constant volume: P₁/T₁ = P₂/T₂
New pressure, P₂ = P₁ × (T₂/T₁) = (6.25 atm) × (373/295) = 7.90 atm
New: P₂ = ? atm, T₂ = (273 + 100) K = 373 K
For a fixed amount of gas at constant volume: P₁/T₁ = P₂/T₂
New pressure, P₂ = P₁ × (T₂/T₁) = (6.25 atm) × (373/295) = 7.90 atm
2018-04-09 10:42 am
If the volume does NOT change, the new pressure is
(6.25 atm)*(373.15K/295.15K) = about 8 atm but use a calculator.
(6.25 atm)*(373.15K/295.15K) = about 8 atm but use a calculator.
收錄日期: 2021-05-01 14:17:30
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