Answer:
∆S1 = 0.5166kJ/K
∆S2 = 0.51826kJ/K
Explanation:
Check attachment for solution
Answer:
Explanation:
Tu pon lo que ocupes y espera
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
What you mean?? .!.!.!.!!.
OA bloom is smaller than a bar
