A sample of nitrogen occupies 10.0 liters at 25°C and 98.7 kPa. What would be the volume at 20°C and 102.7 kPa?
回答 (2)
2017-05-20 12:08 am
Initial: P₁ = 98.7 kPa, V₁ = 10.0 L, T₁ = (273 + 25) K = 298 K
Final: P₂ = 102.7 kPa, V₂ = ? L, T₂ = (273 + 20) K = 293 K
For a fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂
Then, V₂ = V₁ × (P₁/P₂) × (T₂/T₁)
Final volume. V₂ = (10.0 L) × (98.7/102.7) × (293/298) = 9.45 L
Final: P₂ = 102.7 kPa, V₂ = ? L, T₂ = (273 + 20) K = 293 K
For a fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂
Then, V₂ = V₁ × (P₁/P₂) × (T₂/T₁)
Final volume. V₂ = (10.0 L) × (98.7/102.7) × (293/298) = 9.45 L
2017-05-20 12:08 am
You can use P1V1/T1 = P2V2/T2
(10.0 L)(98.7 kPa)/298K = (102.7 kPa)(V2)/293K NOTE: temperature is in degrees K
V2 = 9.45 liters
(10.0 L)(98.7 kPa)/298K = (102.7 kPa)(V2)/293K NOTE: temperature is in degrees K
V2 = 9.45 liters
收錄日期: 2021-04-18 16:56:25
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