Given:
mass of water, m = 2000 kg
temperature, T = 
= 303 K
extacted mass of water = 100 kg
Atmospheric pressure, P = 101.325 kPa
Solution:
a) Using Ideal gas equation:
PV = m
T (1)
where,
V = volume
m = mass of water
P = atmospheric pressure
R= Rydberg's constant = 8.314 KJ/K
M = molar mass of water = 18 g/ mol
Now, using eqn (1):
Therefore, the volume of the tank is
b) After extracting 100 kg of water, amount of water left, m' = m - 100
m' = 2000 - 100 = 1900 kg
The remaining water reaches thermal equilibrium with surrounding temperature at T' = = 303 K
At equilibrium, volume remain same
So,
P'V = m'T'
Therefore, the final pressure is P' = 96.258 kPa
Answer: 100% (double)
Explanation:
The question tells us two important things:
- Mass remains constant
- Volume remains constant
(We can think in a gas enclosed in a closed bottle, which is heated, for instance)
In this case we know that, as always the gas can be considered as ideal, we can apply the general equation for ideal gases, as follows:
- State 1 (P1, V1, n1, T1) ⇒ P1*V1 = n1*R*T1
- State 2 (P2, V2, n2, T2) ⇒ P2*V2 = n2*R*T2
But we know that V1=V2 and that n1=n2, som dividing both sides, we get:
P1/P2 = T1/T2, i.e, if T2=2 T1, in order to keep both sides equal, we need that P2= 2 P1.
This result is just reasonable, because as temperature measures the kinetic energy of the gas molecules, if temperature increases, the kinetic energy will also increase, and consequently, the frequency of collisions of the molecules (which is the pressure) will also increase in the same proportion.
Answer:
1839
Explanation:
the first huge step in solar penal technology came when alxendare becquerel observed the photovolatic effect in 1839,which occurs when a material produces electric current when exposed to light.it was not untill 1888 that the first solar cell was actually built by aleksander stoletov.
Answer:
See explanation
Explanation:
Given:
Initial pressure,
p
1
=
15
psia
Initial temperature,
T
1
=
80
∘
F
Final temperature,
T
2
=
200
∘
F
Find the gas constant and specific heat for carbon dioxide from the Properties Table of Ideal Gases.
R
=
0.04513
Btu/lbm.R
C
v
=
0.158
Btu/lbm.R
Find the work done during the isobaric process.
w
1
−
2
=
p
(
v
2
−
v
1
)
=
R
(
T
2
−
T
1
)
=
0.04513
(
200
−
80
)
w
1
−
2
=
5.4156
Btu/lbm
Find the change in internal energy during process.
Δ
u
1
−
2
=
C
v
(
T
2
−
T
1
)
=
0.158
(
200
−
80
)
=
18.96
Btu/lbm
Find the heat transfer during the process using the first law of thermodynamics.
q
1
−
2
=
w
1
−
2
+
Δ
u
1
−
2
=
5.4156
+
18.96
q
1
−
2
=
24.38
Btu/lbm
