<span>THIS IS A GAS PHASE REACTION AND WE ARE GIVE PARTIAL PRESSURES . I WRITE IN TERMS OF P RATHER THAN CONCENTRATION :
lnPso2cl12=-kt+lnPso2cl1
initial partial pressure Pso2cl12 the rate constant k and the time t
lnPso2cl12=(4.5*10-2*s-1)*65*s+ln (375)
so lnPso2cl12=3.002
we take the base e antilog:
lnPso2cl12=e3.002
Pso2cl12=20 torr
we use the integrated first order rate
lnPso2cl12=3.002=k*t+ lnPso2cl12=3.002
we use the same rate constant and initial pressure
k=4.5*10-2*s-1
Pso2cl12=375
Pso2cl12=1* so2cl12
Pso2cl12=37.5 torr
subtract in Pso2cl12 grom both side
lnPso2cl12- lnPso2cl12=-kt
ln(x)-ln(y)=ln (x/y)
ln (Pso2cl12/Pso2cl20)=-kt
we get t
-1/k*ln(Pso2cl12/Pso2cl20)=t
t=51 s</span>
Answer:
physical cosmology, the Big Rip is a hypothetical cosmological model concerning the ultimate fate of the universe, in which the matter of the universe, from stars and galaxies to atoms and subatomic particles, and even spacetime itself, is progressively torn apart by the expansion of the universe at a certain time
The force exerted between 2 point charges is directly proportional to the product of their strengths, is inversely proportional to the square of the distance between them and is inversely proportional to the absolute permittivity of the surrounding mediums.
I hope that helps somewhat, I'm only kind of sure that could be correct
To solve this problem we will start by differentiating the values in each of the states of matter. Subsequently through the thermodynamic tables we will look for the values related to the entropy, enthalpy and respective specific volumes. Through the relationship of Power defined as the product between mass and enthalpy and mass, specific volume and pressure, we will find the energetic values in the two states investigated. We will start defining the states
State 1
![T_1 = 700\°C](https://tex.z-dn.net/?f=T_1%20%3D%20700%5C%C2%B0C)
![P_1 = 4 Mpa](https://tex.z-dn.net/?f=P_1%20%3D%204%20Mpa)
From steam table
![h_1 =3906.41 KJ/Kg](https://tex.z-dn.net/?f=h_1%20%3D3906.41%20KJ%2FKg)
![s_1 = 7.62 KJ/Kg.K](https://tex.z-dn.net/?f=s_1%20%3D%207.62%20KJ%2FKg.K)
Now
<em>As 1-2 is isentropic</em>
State 2
![P_2 = 20 Kpa](https://tex.z-dn.net/?f=P_2%20%3D%2020%20Kpa)
From steam table
![h_2 = 2513.33 KJ/Kg](https://tex.z-dn.net/?f=h_2%20%3D%202513.33%20KJ%2FKg)
PART A) The power produced by turbine is the product between the mass and the enthalpy difference, then
![Power = m \times (h_1-h_2)](https://tex.z-dn.net/?f=Power%20%3D%20m%20%5Ctimes%20%28h_1-h_2%29)
![P = (50)(3906.41 - 2513.33)](https://tex.z-dn.net/?f=P%20%3D%20%2850%29%283906.41%20-%202513.33%29)
![P = 69654kW](https://tex.z-dn.net/?f=P%20%3D%2069654kW)
b) Pump Work
State 3
![P_3 = 20 Kpa](https://tex.z-dn.net/?f=P_3%20%3D%2020%20Kpa)
![\upsilon= 0.001 m^3/kg](https://tex.z-dn.net/?f=%5Cupsilon%3D%200.001%20m%5E3%2Fkg)
The Work done by the pump is
![W= m\upsilon \Delta P](https://tex.z-dn.net/?f=W%3D%20m%5Cupsilon%20%5CDelta%20P)
![W = (50)(0.001)(4000-20)](https://tex.z-dn.net/?f=W%20%3D%20%2850%29%280.001%29%284000-20%29)
![W = 199kJ](https://tex.z-dn.net/?f=W%20%3D%20199kJ)
Since the 2-watt jobber has to dissipate twice as much power as the smaller unit I would expect it to have double the surface area. So if they are the usual cylindrical striped resistors then the larger one could have either double the length or else 1.4 times the radius of the smaller one.