Which of the following was not a contribution by Niels Bohr?Electrons are in orbits. There is a relationship between the outer s
2 answers:
Answer: The charge of an electron
Explanation:
The charge on electrons was determined by the Robert Millikan and Harvy Fletcher from their experiment named Oil drop experiment.
The charge on the electrons is
Whereas Neil Bohr in his structure of atom explained that:
- Electrons are moving around the nucleus in a discrete circular paths known as orbits or shells.
- These orbitals or shells have fixed value of energy.
- Emission spectra of hydrogen.
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Answer:
(a) Δ = 2.881 J/K; Δ
= -2.881 J/K; total change in entropy = 0
(b)Δ = 2.881 J/K; Δ
= 0 ; total change in entropy = 2.881 J/K
(c) Δ = 0 ; Δ
= 0 ; total change in entropy = 0
Explanation:
In the given problem, we need to calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting total change in entropy, when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume. We have the following variable:
mass (m) = 14 g
Temperature = 298 K
Pressure = 1.00 bar
Initial volume =
Final volume = =
(a) Change in entropy of the system Δ =
where R = 8.314 J/(mol*K)
n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles
Δ = 0.5*8.314*ln2 = 2.881 J/K
Change in entropy of the surrounding Δ = -2.881 J/K
Therefore, for a reversible process, the total change in entropy = Δ+Δ
= 2.881 - 2.881 = 0
(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, Δ = 2.881 J/K
Since surrounding does not change in this process Δ = 0.
total change in entropy = Δ+Δ
= 2.881 - 0 = 2.88 J/K
(c) For an adiabatic reversible expansion, q(rev) = 0, thus:
Δ = 0
Since heat energy is not transferred from the system to the surrounding
Δ = 0
total change in entropy = Δ+Δ
= 0