Which of the following numbers is correctly rounded to three significant figures, as shown in brackets?
1 answer:
You might be interested in
<u>Answer:</u>
The mass of the atom depends on the sub atomic particles present in the nucleus of an atom.
That is, the protons and the neutrons.
Electrons are present around the nucleus and the mass is negligible since its mass is very very less.
<em>mass of a proton = </em>
<em>mass of a neutron = </em>
<em>mass of an electron = </em>
Mass number represents the mass of one particular isotope and it is a whole number for example,
Mass number is 13 and atomic number is 6 for the carbon isotope C-13.
Atomic mass is different from mass number and it is a fraction since it is the average atomic mass of all the isotopes of an atom.
Atomic mass of C is 12.011 amu which we see in the periodic table is the average atomic mass of isotopes C-12, C-13 and C-14.
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