Answer 1 Correct answer is d) 59.9 kJ
Enthalpy change (dH) is related to Internal energy change (dE) and volume change (dV) as
dH = dE + PdV
dE = 58.0 kJ
dV = 12.6 L
PdV = 1.50 atm * 12.6 L = 18.9 L atm
= 18.9 L atm * 101.325 J / 1000 J/kJ
= 1.92 kJ
So, dH = 58.0 kJ + 1.92 kJ
= 59.9 kJ (option d)
Answer 2 Correct answer is option a) 110 kJ
At constant pressure, q = dH
So, dH = 125 kJ
As per first law of thermodynamics:
dE = q + w
= 125 kJ + (-15 kJ)
= 110 kJ (option a)
Share At a constant pressure of 1.50 atm what is the enthalpy change AH for a...
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13. A gaseous reaction occurs in a piston and has an enthalpy change of ?H=253.0 J. Calculate the change in internal energy, ?U, if the volume expands from 1.0 L to 2.0 L against a constant pressure of 2.0 atm. 101.325 J = 1 L?atm. +50.4 J -50.4 J +455.7 J -455.7 J
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