A balloon contains 125.0ml of hydrogen at 631 mm Hg at 36 degrees Celsius. What is the volume of the gas at 417 m Hg and 31 degrees celcius?
回答 (2)
2017-04-25 1:47 pm
Initial: P₁ = 631 mmHg, V₁ = 125.0 ml, T₁ = (273 + 36) K = 309 K
Final: P₂ = 417 mmHg, V₂ = ? ml, T₂ = (273 + 31) K = 304 K
For a fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂
Hence, V₂ = V₁ × (P₁/P₂) × (T₂/T₁)
Final volume, V₂ = (125.0 ml) × (631/417) × (304/309) = 186.1 ml
Final: P₂ = 417 mmHg, V₂ = ? ml, T₂ = (273 + 31) K = 304 K
For a fixed amount of gas: P₁V₁/T₁ = P₂V₂/T₂
Hence, V₂ = V₁ × (P₁/P₂) × (T₂/T₁)
Final volume, V₂ = (125.0 ml) × (631/417) × (304/309) = 186.1 ml
2017-04-25 1:53 pm
The ideal gas law : P1V1/T1 = P2V2/T2 =>
V2 = P1V1T2 / T1P2 = [ 631 mm Hg x 125.0 ml x
(31+ 273 )] / [ ( 36 + 273 ) x 417 mm Hg ] = answer
V2 = P1V1T2 / T1P2 = [ 631 mm Hg x 125.0 ml x
(31+ 273 )] / [ ( 36 + 273 ) x 417 mm Hg ] = answer
收錄日期: 2021-04-18 16:09:08
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