Answer:
The electron stays in an excited state for a short time. When the electron transits from an excited state to its lower energy state, it will gice off the same amound of energy needed to raise to that level. This emitted energy is a photon.
Answer:
P2=0.385atm
Explanation:
step one:
Given that the temperature T1= 60 Celcius
we can convert this to kelvin by adding 273k to 60 Celcius
we have T1= 333k
pressure P1= 0.470 atm
step two:
we know that the standard temperature is T2= 273K
Applying the temperature and pressure relationship we have
P1/T1=P2/T2
substituting our given data we have
0.47/333=P2/273
cross multiply we have
P2= (0.47*273)/333
P2= 128.31/333
P2=0.385 atm
Answer:
ΔU = 103.54 KJ
Explanation:
∴ ΔU = Q + W
ideal gas:
∴ PV = nRT
∴ R = 8.314 L.KPa/K.mol
∴ n = 3000 g N2 * ( mol/28.0134g N2) = 107.143 mol N2
∴ m N2 = 3.0 Kg
∴ T1 = 300 K
∴ P1 = 100 KPa
∴ V1 = nRT1/P1 = 2672.36 L = 2.67 m³
⇒ V2 = 0.9*V1 = 2405.12 L = 2.41 m³
∴ P2 = 140 KPa
⇒ T2 = P2.V2/n.R = 377.99 ≅ 378 K
⇒ W = P1V1 - P2V2
⇒ W = ((100KPa)*(2.67m³)) - ((140KPa)*(2.41m³))
⇒ W = - 70.164 KJ
∴ Q = nCpΔT
∴ Cp = (5/2)*R = 20.785 J/mol.K ....ideal gas
⇒ Q = (107.143mol)*(20.785 J/mol.K)*(378 - 300)
⇒ Q = 173703.446 J = 173.703 KJ
⇒ ΔU = 173.703 KJ - 70.164 KJ
⇒ ΔU = 103.54 KJ
