1 kg of a fluid expands reversibly according to a linear law from 4.2 bar to 1.4 bar; the initial and final volumes are 0.004 m3 and 0.02 m3. The fluid is then cooled reversibly at constant pressure, and finally compressed reversibly according to a law pv = constant back to the initial conditions of 4.2 bar and 0.004 m3. Calculate the work done in each process and the net work of the cycle. Sketch the cycle on a p-v diagram.
please answer me
Answers
Answer:
2890joules
Explanation:
W_{1\to2} = ?;\ W_{2\to3} = ?;\ W_{3\to1} = ?W
1→2
=?; W
2→3
=?; W
3→1
=?
\begin{aligned} W_{1\to2} &= \frac{1}{2}(P_1+P_2)(V_2-V_1)\\ &= \frac{1}{2}((5.2+2.4)×10^5)(0.03-0.005)\\ &= 9500J \end{aligned}
W
1→2
=
2
1
(P
1
+P
2
)(V
2
−V
1
)
=
2
1
((5.2+2.4)×10
5
)(0.03−0.005)
=9500J
\begin{aligned} W_{2\to3} &= P_2(V_3-V_2)\\ \textsf{Accord}&\textsf{ing to Boyles Law}\\ V_3 &= \frac{P_1}{P_3}V_1 = \frac{5.2}{2.4}×0.005 = 0.0108m^3\\ W_{2\to3} &= 2.4×10^5(0.0108-0.03)\\ &= -4608J \end{aligned}
W
2→3
Accord
V
3
W
2→3
=P
2
(V
3
−V
2
)
ing to Boyles Law
=
P
3
P
1
V
1
=
2.4
5.2
×0.005=0.0108m
3
=2.4×10
5
(0.0108−0.03)
=−4608J
\begin{aligned} W_{3\to1} &= P_1V_1\ln\frac{V_1}{V_3}\\ &= 5.2×10^5 × 0.005 × \ln0.463\\ &= -2002J \end{aligned}
W
3→1
=P
1
V
1
ln
V
3
V
1
=5.2×10
5
×0.005×ln0.463
=−2002J
\therefore∴ The net work of the cycle is;
9500 - 4608 - 2002 = 2890J
Answer:
HOPE THIS HELPS YOU
PLS MARK ME AS THE BRIANLIEST
