Answer:
I think it is process or technology
Answer:
74,4 litros
Explanation:
Dado que
W = nRT ln (Vf / Vi)
W = 3000J
R = 8,314 JK-1mol-1
T = 58 + 273 = 331 K
Vf = desconocido
Vi = 25 L
W / nRT = ln (Vf / Vi)
W / nRT = 2.303 log (Vf / Vi)
W / nRT * 1 / 2.303 = log (Vf / Vi)
Vf / Vi = Antilog (W / nRT * 1 / 2.303)
Vf = Antilog (W / nRT * 1 / 2.303) * Vi
Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25
Vf = 74,4 litros
Answer:
Always make sure that you are properly protected and that it is all clear to be operated on.
Answer:
11548KJ/kg
10641KJ/kg
Explanation:
Stagnation enthalpy:
![h_{T} = c_{p}*T + \frac{V^2}{2}](https://tex.z-dn.net/?f=h_%7BT%7D%20%3D%20c_%7Bp%7D%2AT%20%2B%20%5Cfrac%7BV%5E2%7D%7B2%7D)
given:
cp = 1.0 KJ/kg-K
T1 = 25 C +273 = 298 K
V1 = 150 m/s
![h_{1} = (1.0 KJ/kg-K) * (298K) + \frac{150^2}{2} \\\\h_{1} = 11548 KJ / kg](https://tex.z-dn.net/?f=h_%7B1%7D%20%3D%20%281.0%20KJ%2Fkg-K%29%20%2A%20%28298K%29%20%2B%20%5Cfrac%7B150%5E2%7D%7B2%7D%20%5C%5C%5C%5Ch_%7B1%7D%20%3D%20%2011548%20KJ%20%2F%20kg)
Answer: 11548 KJ/kg
Using Heat balance for steady-state system:
![Flow(m) *(h_{1} - h_{2} + \frac{V^2_{1} - V^2_{2} }{2} ) = Q_{in} + W_{out}\\](https://tex.z-dn.net/?f=Flow%28m%29%20%2A%28h_%7B1%7D%20-%20h_%7B2%7D%20%2B%20%5Cfrac%7BV%5E2_%7B1%7D%20-%20V%5E2_%7B2%7D%20%20%7D%7B2%7D%20%29%20%3D%20Q_%7Bin%7D%20%2B%20W_%7Bout%7D%5C%5C)
Qin = 42 MW
W = -100 KW
V2 = 400 m/s
Using the above equation
![50 *( 11548- h_{2} + \frac{150^2 - 400^2 }{2} ) = 42,000 - 100\\\\h_{2} = 10641KJ/kg](https://tex.z-dn.net/?f=50%20%2A%28%2011548-%20h_%7B2%7D%20%2B%20%5Cfrac%7B150%5E2%20-%20400%5E2%20%7D%7B2%7D%20%29%20%3D%2042%2C000%20-%20100%5C%5C%5C%5Ch_%7B2%7D%20%3D%2010641KJ%2Fkg)
Answer: 10641 KJ/kg
c) We use cp because the work is done per constant pressure on the system.
Answer:
<em>Never</em>
Explanation:
In a reversible adiabatic process, there is not transfer of heat or matter between the system and its environment. An adiabatic reversible process is a process with constant entropy, i.e ΔQ=0. The internal energy is solely dependent on the work done either due to compression or expansion. So the entropy of the gas will never increase.