Explanation:
Hello,
In this case, given the enthalpies and entropies of boh fusion and boiling, we can compute both the fusion and boiling points as shown below, considering ethanol's molar mass is 46 g/mol:
Best regards.
Explanation:
Hello,
In this case, given the enthalpies and entropies of boh fusion and boiling, we can compute both the fusion and boiling points as shown below, considering ethanol's molar mass is 46 g/mol:
Best regards.
A) 251.8 kj/kg
B) 0.9150 kj/kg-k
C) 155.4 kj/kg
F) 1.50
G) 3.95 kw
H) 2.6 kw
Explanation:
Given conditions :
air conditioner : R -134a
compressor efficiency (nc) = 90%.
T1 = 9⁰c, T3 = 39⁰c, mass flow rate = 0.027 kg/s
A) Specific enthalpy at the compressor inlet
at T = 9⁰c the saturated vapor (x) = 1
from the R-134a property table
h1 = 251.8 kj/kg
B ) specific entropy ( kj/kg-k) at the compressor inlet
at T = 9⁰c the saturated vapor (x) = 1
s = 0.9150 kj/kg-k ( from the R-134a property table )
C) specific enthalpy at the compressor exit
at T3 = 39⁰c , s2 = s1
has = 165.12 kj/kg
h2 = 155.4 kj/kg
attached below is the remaining solution to some of the problems
A) 251.8 kj/kg
B) 0.9150 kj/kg-k
C) 155.4 kj/kg
F) 1.50
G) 3.95 kw
H) 2.6 kw
Explanation:
Given conditions :
air conditioner : R -134a
compressor efficiency (nc) = 90%.
T1 = 9⁰c, T3 = 39⁰c, mass flow rate = 0.027 kg/s
A) Specific enthalpy at the compressor inlet
at T = 9⁰c the saturated vapor (x) = 1
from the R-134a property table
h1 = 251.8 kj/kg
B ) specific entropy ( kj/kg-k) at the compressor inlet
at T = 9⁰c the saturated vapor (x) = 1
s = 0.9150 kj/kg-k ( from the R-134a property table )
C) specific enthalpy at the compressor exit
at T3 = 39⁰c , s2 = s1
has = 165.12 kj/kg
h2 = 155.4 kj/kg
attached below is the remaining solution to some of the problems
Explanation:
First, we will calculate the entropies as follow.
J/K mol
As, = 298 K, = 373.8 K
Putting the given values into the above formula we get,
J/K mol
=
= 18.5 J/K mol
Now,
=
= 94.2 J/mol K
Also,
=
= 33.4 J/K mol
Now, we will calculate the entropy of one mole of methanol vapor at 800 K as follows.
= (126.8 + 18.5 + 94.2 + 33.4) J/K mol
= 272.9 J/K mol
Thus, we can conclude that the entropy of one mole of methanol vapor at 800 K is 272.9 J/K mol.
Explanation:
First, we will calculate the entropies as follow.
J/K mol
As, = 298 K, = 373.8 K
Putting the given values into the above formula we get,
J/K mol
=
= 18.5 J/K mol
Now,
=
= 94.2 J/mol K
Also,
=
= 33.4 J/K mol
Now, we will calculate the entropy of one mole of methanol vapor at 800 K as follows.
= (126.8 + 18.5 + 94.2 + 33.4) J/K mol
= 272.9 J/K mol
Thus, we can conclude that the entropy of one mole of methanol vapor at 800 K is 272.9 J/K mol.
a. ΔG : It seems that the proper definition is not given, anyway, it is the free energy change.
b. ΔS : 3. Change in entropy.
c. Keq : 6. Equilibrium constant.
d. ΔH : 7. Change in enthalpy.
e. ΔG° : 9. Standard free energy change.
f. J : 8. Joule.
Explanation:
Hello,
In this case, we can match the symbol with the proper definition as shown below:
a. ΔG : It seems that the proper definition is not given, anyway, it is the free energy change and it uses G since it is better referred to the Gibbs free energy.
b. ΔS : 3. Change in entropy.
c. Keq : 6. Equilibrium constant.
d. ΔH : 7. Change in enthalpy.
e. ΔG° : 9. Standard free energy change.
f. J : 8. Joule.
Best regards.
ΔS = -0.272 J/K
Explanation:
Step 1: Data given
Its normal freezing point is –38.9 °C
molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol
Mass of Hg = 5.590 grams
Step 2:
ΔG = ΔH - TΔS
At the normal freezing point, or any phase change in general,
ΔG =0
0 = Δ
Hfus
−
Tfus Δ
S
fus
Δ
S
fus = Δ
Hfus
/Tfus
Δ
S
fus = 2290 J/mol / 234.25 K
Δ
S
fus = 9.776 J/mol*K
Since fusion is from solid to liquid. Freezing is the opposite process, so the entropy change of freezing is -9.776 J/mol*K
Step 3: Calculate moles Hg
Moles Hg = 5.590 grams / 200.59 g/mol
Moles Hg = 0.02787 moles
Step 4: Calculate the entropy change of the system:
Δ
S = -9.776 J/mol*K * 0.02787 moles
ΔS = -0.272 J/K
c. The statement is true
Explanation:
Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and for the reaction comes out to be positive.
Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and for the reaction comes out to be negative.
As the given value of , the heat is released and is exothermic.Thus the statement that the reaction is exothermic is true.
ΔS = -0.272 J/K
Explanation:
Step 1: Data given
Its normal freezing point is –38.9 °C
molar enthalpy of fusion is ∆Hfusion = 2.29 kJ/mol
Mass of Hg = 5.590 grams
Step 2:
ΔG = ΔH - TΔS
At the normal freezing point, or any phase change in general,
ΔG =0
0 = Δ
Hfus
−
Tfus Δ
S
fus
Δ
S
fus = Δ
Hfus
/Tfus
Δ
S
fus = 2290 J/mol / 234.25 K
Δ
S
fus = 9.776 J/mol*K
Since fusion is from solid to liquid. Freezing is the opposite process, so the entropy change of freezing is -9.776 J/mol*K
Step 3: Calculate moles Hg
Moles Hg = 5.590 grams / 200.59 g/mol
Moles Hg = 0.02787 moles
Step 4: Calculate the entropy change of the system:
Δ
S = -9.776 J/mol*K * 0.02787 moles
ΔS = -0.272 J/K
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