Answer:
learn from their mistakes
Explanation:
and so u could do better next time
Answer:
-6.326 KJ/K
Explanation:
A) the entropy change is defined as:
![delta S_{12}=\int\limits^2_1 \, \frac{dQ}{T}](https://tex.z-dn.net/?f=delta%20S_%7B12%7D%3D%5Cint%5Climits%5E2_1%20%20%5C%2C%20%5Cfrac%7BdQ%7D%7BT%7D)
In an isobaric process heat (Q) is defined as:
![Q= m*Cp*dT](https://tex.z-dn.net/?f=Q%3D%20m%2ACp%2AdT)
Replacing in the equation for entropy
m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:
Solving the integral we get the expression to estimate the entropy change in the system
![delta S_{12}= m*Cp *ln(\frac{T_{2}}{T_{1}})](https://tex.z-dn.net/?f=delta%20S_%7B12%7D%3D%20m%2ACp%20%2Aln%28%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D%29)
The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is ![0.85\frac{kJ}{Kg*K}](https://tex.z-dn.net/?f=0.85%5Cfrac%7BkJ%7D%7BKg%2AK%7D)
We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K
The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.
![Q= m*Cp*dT](https://tex.z-dn.net/?f=Q%3D%20m%2ACp%2AdT)
With
clearing for T2 we get:
![T_{2}=\frac{Q}{m*Cp}+T1= \frac{-976.71kJ}{5.25Kg*0.85\frac{kJ}{Kg*K}}+288.86 K =69.98 K](https://tex.z-dn.net/?f=T_%7B2%7D%3D%5Cfrac%7BQ%7D%7Bm%2ACp%7D%2BT1%3D%20%5Cfrac%7B-976.71kJ%7D%7B5.25Kg%2A0.85%5Cfrac%7BkJ%7D%7BKg%2AK%7D%7D%2B288.86%20K%20%3D69.98%20K)
Now we can estimate the entropy change in the system
![delta S_{12}= m*Cp*ln(\frac{T_{2}}{T_{1}})= 5.25Kg*0.85\frac{kJ}{Kg*K}*ln(\frac{69.98}{288.861})= -6.326\frac{kJ}{K}](https://tex.z-dn.net/?f=delta%20S_%7B12%7D%3D%20m%2ACp%2Aln%28%5Cfrac%7BT_%7B2%7D%7D%7BT_%7B1%7D%7D%29%3D%205.25Kg%2A0.85%5Cfrac%7BkJ%7D%7BKg%2AK%7D%2Aln%28%5Cfrac%7B69.98%7D%7B288.861%7D%29%3D%20-6.326%5Cfrac%7BkJ%7D%7BK%7D)
The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.
b) see picture.
The one that is not an option of the 3 technology bets made are Digital core and Design Thinking.
<h3>What are the 3 technology bets Genpact produced?</h3>
The digital technologies made are known to be able to create value through the accelerating processes and also by automating them.
The technology bets Genpact are:
- Artificial Intelligence.
- Augmented Intelligence.
- Customer Experience.
- Digital Transformation and AI Consulting.
- Intelligent Automation.
Learn more about technology from
brainly.com/question/25110079
Answer: Option D) 298 g/mol is the correct answer
Explanation:
Given that;
Mass of sample m = 13.7 g
pressure P = 2.01 atm
Volume V = 0.750 L
Temperature T = 399 K
Now taking a look at the ideal gas equation
PV = nRT
we solve for n
n = PV/RT
now we substitute
n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K
)
= 1.5075 / 32.7579
= 0.04601 mol
we know that
molar mass of the compound = mass / moles
so
Molar Mass = 13.7 g / 0.04601 mol
= 297.7 g/mol ≈ 298 g/mol
Therefore Option D) 298 g/mol is the correct answer