Answer: Total work produced = 1.355 × 10^5 KJ
Explanation:
Substance :H2O, V1=1m3
State1= sat. liquid, T1=200°C, x1=0
State2= sat.mixture, T1=200°C, x2= 0.8
v1=vf=0.001157m3/kg, vg=0.12721m3/kg
P1=P2=Psat. =1554.9 KPa
v2= vf+x2 (vg-vf) =0.001157 +0.8(0.12721-0.001157)=0.102m3/kg
V2= V1×v2/v1 = 1× 0.102/0.001157 =88.159m3
W= P(V1-V2) =1554.9(88.152- 1) = 1.355x10^5KJ
Answer:
a) Mat swum 100 meters.
b)after 4 length , Mat's displacement is zero.
Explanation:
As given, Mat and David are training for the swimming team , and Length of 1 side of the pool is 25 meter.
As swimming pool will be a cube or a cuboid , so it has only 4 sides.
a)
1 length = 23 meter
⇒4 length = 4×25 = 100 meter
So, Mat swam 100 meters.
b)
As after swam 4 lengths of the pool, Mat reaches to its initial position i.e Mat come back to its starting point.
As Displacement is always measures from the starting point.
So, after 4 length , Mat's displacement is zero.
Complete Question
The complete question is shown on the first uploaded image
Answer:
Explanation:
From the question we are told that
The initial pressure of the gas is ![P_1 = 1.50 \ atm](https://tex.z-dn.net/?f=P_1%20%3D%20%201.50%20%5C%20%20atm)
The volume of the second bulb is ![V_2 = 0.800 \ L](https://tex.z-dn.net/?f=V_2%20%3D%20%200.800%20%5C%20%20L)
The final pressure of the gas is ![P_2 = 671 \ torr = \frac{671 }{ 760} = 0.883 \ atm](https://tex.z-dn.net/?f=P_2%20%3D%20%20671%20%5C%20torr%20%3D%20%20%5Cfrac%7B671%20%7D%7B%20760%7D%20%20%3D%20%200.883%20%5C%20atm)
Let the unknown volume be represented as
Generally from Boyle's law we have that
![P_1 * V_1 = P_2 * V_2](https://tex.z-dn.net/?f=P_1%20%2A%20%20V_1%20%3D%20%20P_2%20%2A%20V_2)
=> ![1.50 * V = 0.883 * 0.800](https://tex.z-dn.net/?f=1.50%20%2A%20%20V%20%3D%20%20%200.883%20%20%2A%200.800)
=> ![V = 0.479 \ L](https://tex.z-dn.net/?f=V%20%20%3D%200.479%20%5C%20L)
The answer is around 7.
As the acid and the base are of equal strength, they neutralize each other and the resulting solution is neutral (pH≈7).
Hope this helps you!
(5 L/m)x(1,440 m/da)x(365 da/yr)x(20yr)
That's 52,560,000 Liters, or 52,560 m^3.