Answer:
V₂ = 107.84 L
Explanation:
Given data:
Initial volume = 100 L
Initial pressure = 80 KPa (80/101 =0.79 atm)
Initial temperature = 200 K
Final temperature =273 K
Final volume = ?
Final pressure = 1 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁T₂ /T₁P₂
V₂ = 0.79 atm × 100 L × 273 K / 200 K × 1 atm
V₂ =21567 atm.L.K /200 K.atm
V₂ = 107.84 L
Answer:
1) ΔG°r(298 K) = - 28.619 KJ/mol
2) ΔG°r will decrease with decreasing temperature
Explanation:
- CO(g) + H2O(g) → H2(g) + CO2(g)
1) ΔG°r = ∑νiΔG°f,i
⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)
from literature, T = 298 K:
∴ ΔG°CO2(g) = - 394.359 KJ/mol
∴ ΔG°CO(g) = - 137.152 KJ/mol
∴ ΔG°H2(g) = 0 KJ/mol........pure substance
∴ ΔG°H2O(g) = - 228.588 KJ/mol
⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )
⇒ ΔG°r(298 K) = - 28.619 KJ/mol
2) K = e∧(-ΔG°/RT)
∴ R = 8.314 E-3 KJ/K.mol
∴ T = 298 K
⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6
⇒ ΔG°r = - RTLnK
If T (↓) ⇒ ΔG°r (↓)
assuming T = 200 K
⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)
⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol
No He believed tiny particles were invisible and couldn't be changed....So No The person that believed in this was Dalton .
Answer: The answer is D :)
Explanation:
Answer:
0.33 cal⋅g-1°C-1
Explanation:
The amount of heat required is determined from the formula:
q= mcΔT
To see more:
https://api-project-1022638073839.appspot.com/questions/what-is-the-specific-heat-of-a-substance-if-1560-cal-are-required-to-raise-the-t#235434
