更新1:
at 100 ∘C? ΔHcombustion=−8.90×102kJmol−1CH4. For H2O(l) at 100 ∘C, d=0.958gcm−3 and ΔHvap=40.7kJmol−1.
How many liters of CH4(g), measured at 23.0 ∘C and 764 mmHg , must be burned to provide the heat needed to vaporize 3.86 L of water a...?
回答 (1)
2016-07-29 6:00 pm
Molar mass of H₂O (water) = (1.006×2 + 16.00) g mol⁻¹ = 18.02 g mol⁻¹
Molar mass of CH₄ = (12.01 + 1.008×4) g mol⁻¹ = 16.04 g mol⁻¹
Heat absorbed by water (in kJ) = [(3860 cm³) × (0.958 g cm⁻³) / (18.02 g mol⁻¹)] × (40.7 kJ mol⁻¹)
Heat released in burning CH₄ (in kJ) = [n(CH₄) mol] × (8.90 × 10² kJ mol⁻¹)
Heat released in burning CH₄ = Heat absorbed by water
n(CH₄) × (8.90 × 10²) = [3860 × 0.958 / 18.02] × 40.7
n(CH₄) = 9.38 mol
For the CH₄ gas :
P = 764 mmHg
V = ? L
n = 9.38 mol
R = 62.36 mmHg L mol⁻¹ K⁻¹
T = (273.2 + 23.0) K = 296.2 K
PV = nRT
V = nRT/P
Volume of CH₄ burned = 9.38 × 62.36 × 296.2 / 764 = 227 L
Molar mass of CH₄ = (12.01 + 1.008×4) g mol⁻¹ = 16.04 g mol⁻¹
Heat absorbed by water (in kJ) = [(3860 cm³) × (0.958 g cm⁻³) / (18.02 g mol⁻¹)] × (40.7 kJ mol⁻¹)
Heat released in burning CH₄ (in kJ) = [n(CH₄) mol] × (8.90 × 10² kJ mol⁻¹)
Heat released in burning CH₄ = Heat absorbed by water
n(CH₄) × (8.90 × 10²) = [3860 × 0.958 / 18.02] × 40.7
n(CH₄) = 9.38 mol
For the CH₄ gas :
P = 764 mmHg
V = ? L
n = 9.38 mol
R = 62.36 mmHg L mol⁻¹ K⁻¹
T = (273.2 + 23.0) K = 296.2 K
PV = nRT
V = nRT/P
Volume of CH₄ burned = 9.38 × 62.36 × 296.2 / 764 = 227 L
收錄日期: 2021-04-20 16:26:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160729083424AAopsxc