The volume of an ideal gas is decreased from 5.0 l to 5.0 ml at a constant pressure of 2.0 atm. calculate the work associated wi
2 answers:
Answer:
The work is 9.99J.
Explanation:
The work associated with a process is given by the following formula:
In which W is the work, P is the pressure and is the difference of volume.
In this problem, we have that:
The volume of an ideal gas is decreased from 5.0 l to 5.0 ml. Each ml has 0.001l. So the volume of the ideal gas is decreased from 5.0l to 0.005l. So
.
Constant pressure of 2.0 atm. So .
The work is 9.99J.
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Answer:
W(r,out) = 5.81 MW
= 86.1 %
Explanation:
we use here steam table for get value of h1, s1 etc
so use for 6MPa and 600 degree
Enthalphy of steam h1 = 3658.8 kJ/kg
Entropy of steam s 1 is = 7.1693 kJ /kg.K
and
for 50 kPa and 100 degree
Enthalphy of steam h2 = 2682.4 kJ/kg
Entropy of steam s2 is = 7.6953 kJ /kg.K
so we use here energy balance equation that is
..............1
put here value and we get m
m =
solve it we get
m = 5.156 kg/s
so by energy balance equation
m1 = m
2 + W(r,out)
W(r,out) = m(1 -
2)
W(r,out) = h1 - h2 + ΔKE + ΔPE - To(s1-s2)
W(r,out) =
W(r,out) = W(a,out) - m.To.(s1-s2) ........................2
put here value
W(r,out) = 5000 - ( 5.156 × (25 + 273) ×( 7.1693 - 7.6953)
W(r,out) = 5908.19 = 5.81 MW
and
second law deficiency is
=
..............................3
put here value
= \frac{5}{5.81}
= 86.1 %