What is the entropy change for the freezing process of 1 mole of liquid methanol at its freezing temperature (–97. 6˚c) and 1 atm? please report your answer two points past the decimal with the unit j/molk. ∆h˚fus = 3. 17 kj/mol.
Answers
Answer:
the value of change in entropy for freezing process of 1 mole of liquid methanol is -18.07 J/mol.k
Explanation:
Entropy is defined as the measure of possible arrangement of the atoms or molecules. Entropy change for the the freezing process mole of 1 liquid methanol at -97.6 degree celcius and 1 atm will be -18.07 J/mol.
The entropy change can be calculated by the given formula:
Thus, the value of change in entropy for freezing process of 1 mole of liquid methanol is -18.07 J/mol.k
(Hope this helps can I pls have brainlist (crown)☺️)
For the freezing of 1 mole of liquid methanol, the change in entropy value is -18.07 J/mol.
What is entropy value?The measure of atoms' or molecules' potential arrangements is known as entropy. Entropy change for one mole of liquid methanol frozen at -97.6 degrees Celsius and one atmosphere is -18.07 J/mol.
The range of entropy is 0 to 1. Depending on the number of classes in your dataset, entropy can be greater than 1, but it still indicates a very high level of chaos. For the sake of convenience, the examples in this blog will have entropy values between 0 and 1.
The "information" a variable contains is measured by its entropy. Now, based on the surprise that this value of the bholds and the number of distinct values that are present in the variable, this amount is approximated.
Thus, it is -18.07 J/mol.
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The question is incomplete. Your most probably complete question is given below:
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Answer:
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Explanation:
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Using implicit differentiation, and considering that the inflation rate is of 200 cubic feet per minute and the radius is of 5 feet, it is found that:
The radius is increasing at a rate of 0.6366 feet per minute.The surface area is changing at a rate of 80 feet squared per minute.The volume of an sphere of radius r is given by:
[tex]V = \frac{4\pi r^3}{3}[/tex]
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[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
For this problem:
Inflation rate of 200 cubic feet per minute, hence [tex]\frac{dV}{dt} = 200[/tex]Radius is of 5 feet, hence [tex]r = 5[/tex]Then:
[tex]\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}[/tex]
[tex]200 = 4\pi (5)^2\frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt} = \frac{200}{100\pi}[/tex]
[tex]\frac{dr}{dt} = 0.6366[/tex]
The radius is increasing at a rate of 0.6366 feet per minute.
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[tex]S = 4\pi r^2[/tex]
Hence, the rate of change is:
[tex]\frac{dS}{dt} = 8\pi r\frac{dr}{dt}[/tex]
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[tex]\frac{dS}{dt} = 8\pi (5)(0.6366) = 80[/tex]
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Answer:
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Explanation:
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Answers
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