two moles of an ideal gas undergoes following
14
process as
A (T.)
Ē 106
(atm)
T. =
ė 0.5
C (T)
40 L
20 L
V(L)
Correct option(s) among the following is/are
(1) AU = AS = 0
(2) Wec=0
(3) Woycle #0
4) All of these
Answers
Answered by
1
Explanation:
AB-Isobaric process
BC-Isochoric process
CA-Isothermal process
Total work =w
AB
+w
BC
+w
CA
=−P×ΔV+2.303nRTlog(
V
1
V
2
)
=−1×20×101.3+0+2.303×2×8.314×Tlog
20
40
...(i)
PV=nRT (At A)
1×20=2×0.0821×T
T=121.8K
From eq.(i) in down given
Total work=−2026+2.303×2×8.314×121.8log2=−622.06J
In cyclic process:
ΔU=0,ΔH=0,ΔS=0
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