Answer:
the entropy change of the fluid during the process process is is 1.337 kJ/K, the change for the source is -1.337 kJ/K and the total entropy change is 0
Explanation:
since the Carnot cycle is a reversible cycle, the entropy change is related with the heat exchanged through:
ΔS =∫dQ/T
since the temperature remains constant
ΔS =∫dQ/T=(1/T)*∫dQ = Q/T
Q= heat added to the system
T= absolute temperature = 400°C= 673 K
therefore
ΔS = Q/T = 900 kJ/ 673 K = 1.337 kJ/K
ΔS working fluid = 1.337 kJ/K
since the process is reversible, the entropy change of the universe (total entropy change) is 0 (there is no entropy generation). thus
ΔS universe = ΔS working fluid + ΔS source = 0
ΔS source= -ΔS working fluid = -1.337 kJ/K
Sp2 hybridization forms 1 sigma bond and 1 pi bond.
<u>Answer:</u> The increase in pressure is 0.003 atm
<u>Explanation:</u>
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= final pressure = ?
= Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![801^oC=[801+273]K=1074K](https://tex.z-dn.net/?f=801%5EoC%3D%5B801%2B273%5DK%3D1074K)
= final temperature = ![(801+1.00)^oC=802.00=[802+273]K=1075K](https://tex.z-dn.net/?f=%28801%2B1.00%29%5EoC%3D802.00%3D%5B802%2B273%5DK%3D1075K)
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B28800J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B1074%7D-%5Cfrac%7B1%7D%7B1075%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D3%5Ctimes%2010%5E%7B-3%7Datm%5C%5C%5C%5CP_2%3De%5E%7B3%5Ctimes%2010%5E%7B-3%7D%7D%3D1.003atm)
Change in pressure = 
Hence, the increase in pressure is 0.003 atm
Answer:
C, when the ball lands it will have the least amount potential energy.
Explanation:
here is a picture that should help you.
Because the Potassium ion loses an electron, the electrons come a little closer to the nucleus because they are attracted to the protons. This makes the ion smaller in radius
