Answer:
- 1 mole of carbon disulfide gas at 273 K and 40 L
- 1 mol of chlorine gas at 273 K and 40 L
- 1 mol of neon gas at 273 K and 40 L
- 1 mol of neon gas at 273 K and 20 L
- 1/2 mol of neon gas at 273 K and 20 L
- 1/2 mol of neon gas at 100 K and 20 L
- 1/2 mol of liquid neon at 100 K
Explanation:
Entropy is the measure of disorder or randomness in a closed system. Its an extensive property of a thermodynamic system
The following points must be considered when ranking the systems according to their entropy:
- The entropy of gases are highest than liquids or solid. And entropy of liquid is higher than solid. That is because gas has more microstate thus have the highest entropy.
- Entropies of large complicated molecules are greater than those of smaller, simpler molecules. Because larger molecules have more disorder because of the greater number of ways they can be move around in three dimensional space.
- highest temperature and highest volume will lead to greatest entropy
- 1 mole of any substance will have greater entropy than 1/2 mole of that same substance
<span>0.967 x 10^-6 HZ
This should be correct (:</span>
Talk to them and listen to each other. if they aren’t ready to talk, give them space. once both of you are ready, you can make up and forgive each other. don’t bother them by asking a lot of questions and forcing them to talk to you. and if they’re doing that, tell them you need time to think. just be sure to talk to them, listen, and understand. tell each other both sides of the stories. of course, different situations can require different solutions. so resolve it when it’s time :)
<h3>Types of Osmosis</h3>
Osmosis is of two types:
Endosmosis– When a substance is placed in a hypotonic solution, the solvent molecules move inside the cell and the cell becomes turgid or undergoes deplasmolysis. This is known as endosmosis.
Exosmosis– When a substance is placed in a hypertonic solution, the solvent molecules move outside the cell and the cell becomes flaccid or undergoes plasmolysis. This is known as exosmosis.
Answer:
There are 3 steps of this problem.
Explanation:
Step 1.
Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.
Step 2.
Enthalpy of saturated liquid Haq = 781.124 J/g
Enthalpy of saturated vapour Hvap = 2779.7 J/g
Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g
Step 3.
In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy
So, H1=H2
H2= (1-x)Haq+XHvap.........1
Putting the values in 1
2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}
= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)
1904.976 (J/g) = x1998.576 (J/g)
x = 1904.976 (J/g)/1998.576 (J/g)
x = 0.953
So, the quality of the wet steam is 0.953
