Which has the greater density? air at sea level air at 20 km altitude
2 answers:
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Question:
Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.
Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.
Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.
Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.
Answer:
D) 85 J/K
E) - 50 J/K
F) 62.5 J/K
G) 12.5 J/K
Explanation:
Let's make use of the entropy equation: ΔS =
Part D)
Given:
T = 20°C = 20 +273 = 293K
Q = 25.0 kJ
Entropy change will be:
ΔS =
= 85 J/K
Part E)
Given:
T = 500K
Q = -25.0 kJ
Entropy change will be:
ΔS =
= - 50 J/K
Part F)
Given:
T = 400K
Q = 25.0 kJ
Entropy change will be:
ΔS =
= 62.5 J/K
Part G:
Given:
T1 = 400K
T2 = 500K
Q = 25.0 kJ
The net entropy change will be:
ΔS =
= 12.5 J/K
The question is incomplete, complete question is;
Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction:
Use the reactions and associated equilibrium constants at body temperature to find the equilibrium constant for the above reaction.
Answer:
The equilibrium constant for the given reaction is 170.
Explanation:
..[1]
..[2]
..[3]
Using [1] in [2]:
( using [3])
The equilibrium constant for the given reaction is 170.