An ideal diatomic gas at 80 k is slowly compressed adiabatically and reversibly to half its volume. The final temperature is 104.8k
<h3>Calculation of final temperature </h3>
The formula used for compression is:-
TV^(γ-1)=C
where;
T= temperature=80k
V=volume(given volume is half of its original volume i.e v/2)
γ=CP/CV= ( 7.R/2)/(5R/2)=7/5
C= constant
using the given values in the formula;
80 x V^[(7/5)-1]=T(final) (V/2)^[(7/5)-1]
80=T(final) x (V/V2)^(2/5)
80=T(final) x (1/2)^(2/5)
T(final)= 80 x (2)^2/5
T(final)= 80 x 1.31
final temperature =104.8k
Learn more about Temperature here:-
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Answer:
a. 599 cm³
b. 1.49 g/cm³
Explanation:
<h3>A. Volume </h3>
Volume is the amount of space an object occupies. Since this is a brick, the object is a rectangular prism. The formula for the volume of a rectangular prism is the product of length, width, and height.
![V= l *w*h](https://tex.z-dn.net/?f=V%3D%20l%20%2Aw%2Ah)
The brick's length (l) is 13.77 centimeters, the width (w) is 8.50 centimeters, and the height (h) is 5.12 centimeters. Substitute these values into the formula.
![V= 13.77 \ cm * 8.50 \ cm * 5.12 \ cm](https://tex.z-dn.net/?f=V%3D%2013.77%20%5C%20cm%20%2A%208.50%20%5C%20cm%20%2A%205.12%20%5C%20cm)
Multiply the numbers together.
![V= 117.045 \ cm^ 2* 5.12 \ cm](https://tex.z-dn.net/?f=V%3D%20117.045%20%5C%20cm%5E%202%2A%205.12%20%5C%20cm)
![V= 599.2704 \ cm^3](https://tex.z-dn.net/?f=V%3D%20599.2704%20%5C%20cm%5E3)
The original measurements have at least 3 significant figures, so our answer must have 3. For the number we calculated, that is the ones place. The 2 in the tenths place tells us to leave the 9 in the ones place.
![V \approx 599 \ cm^3](https://tex.z-dn.net/?f=V%20%5Capprox%20599%20%5C%20cm%5E3)
![\bold {The \ volume \ of \ the \ brick \ is \ approximately \ 599 \ cubic \ centimeters}}](https://tex.z-dn.net/?f=%5Cbold%20%7BThe%20%5C%20volume%20%5C%20of%20%5C%20the%20%5C%20brick%20%5C%20is%20%5C%20approximately%20%5C%20599%20%5C%20cubic%20%5C%20centimeters%7D%7D)
<h3>2. Density </h3>
Density is the amount of matter in a specified space. The formula for density is mass over volume.
![d= \frac{m}{v}](https://tex.z-dn.net/?f=d%3D%20%5Cfrac%7Bm%7D%7Bv%7D)
The mass of the brick is 895.3 grams and we just found the volume to be 599.2704 cubic centimeters. Substitute the values into the formula.
![d= \frac{895.3 \ g}{599 \ cm^3}](https://tex.z-dn.net/?f=d%3D%20%5Cfrac%7B895.3%20%5C%20g%7D%7B599%20%5C%20cm%5E3%7D)
Divide.
![d= 1.494657763 \ g/cm^3](https://tex.z-dn.net/?f=d%3D%201.494657763%20%5C%20g%2Fcm%5E3)
Round to three significant figures. For the number we calculated, that is the hundredth place. The 4 in the thousandth place tells us to leave the 9 in the hundredth place.
![d \approx 1.49 \ g/cm^3](https://tex.z-dn.net/?f=d%20%5Capprox%201.49%20%5C%20g%2Fcm%5E3)
![\bold {The \ density\ of \ the \ brick \ is \ approximately \ 1.49 \ grams /cubic \ centimeters}}](https://tex.z-dn.net/?f=%5Cbold%20%7BThe%20%5C%20density%5C%20of%20%5C%20the%20%5C%20brick%20%5C%20is%20%5C%20approximately%20%5C%201.49%20%5C%20grams%20%2Fcubic%20%5C%20centimeters%7D%7D)
Answer:
D.) Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.
Explanation:
For the reaction:
3H₂(g) + N₂(g) → 2NH₃(g)
The enthalpy change is ΔH = -92kJ
This enthalpy change is defined as the enthalpy of products - the enthalpy of reactants. As the enthalpy is <0, The enthalpy of products is <em>lower </em>than the enthalpy of reactants.
Also, it is possible to obtain the enthalpy change from the bond energies of products - bond energies of reactants, thus, The total bond energies of products are <em>lower</em> than the total bond energies of reactants.
The rate of the reaction couldn't be determined using ΔH.
As the bond energy of ammonia is lower than bonds of nitrogen and hydrogen, <em>D. Nitrogen and Hydrogen are very stable bonds compared to the bonds of ammonia.</em>
I hope it helps!
Is this a multiple choice question? if it is you can show me the answers and then I can figure it out.