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Anon25 [30]
3 years ago
6

Deforestation typically

Chemistry
1 answer:
harkovskaia [24]3 years ago
8 0
<span>B. increases erosion in an area</span>
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Calculate the change in the entropy of the system and also the change in the entropy of the surroundings, and the resulting tota
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(a) ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = -2.881 J/K; total change in entropy = 0

(b)ΔS_{sys}  = 2.881 J/K; ΔS_{sur}  = 0 ; total change in entropy = 2.881 J/K

(c) ΔS_{sys}  = 0 ; ΔS_{sur}  = 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 = V_{1}

Final volume = V_{2} = 2V_{1}

(a) Change in entropy of the system ΔS_{sys} = nRIn\frac{V_{2} }{V_{1} }

where R = 8.314 J/(mol*K)

n = number of moles = mass/molar mass = 14/ 28 = 0.5 moles

ΔS_{sys} = 0.5*8.314*ln2 = 2.881 J/K

Change in entropy of the surrounding ΔS_{sur} = -2.881 J/K

Therefore, for a reversible process, the total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 2.881 = 0

(b) Because entropy is a state function, we use the same procedure as in part (a). Thus, ΔS_{sys}  = 2.881 J/K

Since surrounding does not change in this process ΔS_{sur} = 0.

total change in entropy = ΔS_{sys}+ΔS_{sur} = 2.881 - 0 = 2.88 J/K

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ΔS_{sys}  = 0

Since heat energy is not transferred from the system to the surrounding

ΔS_{sur}  = 0

total change in entropy = ΔS_{sys}+ΔS_{sur} = 0

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