In a reversible isobaric heating process in a closed systemChoose one answer in each lineW12 = p(v1 – vz) OW12 = 0W12 = p(v2 – v1)W12 = v(P1 – P2) O912 = U2 – U1912 = h2 – h0912 = T(S2 - S1) O%3D912 = 0S2 - S1 = 0 OS2- S1 > 00> Is – 2s In a reversible isobaric cooling process in an open system:Choose one answer in each lineW12 = p(v1 – v2) OW12 = 0W12 = p(v2 – v,) O%3DW12 = v(p1 – P2) O%3D912 = u2 – U1912 = h2 – h,O- h,0912 = T(S2 – s1) O%3D912 =0%3DS2 - S1 <0S2 - S1 = 0 OS2 - S1 > 0

In a reversible isobaric process in a closed system, In a closed sytem, there is no transfer of…

In a reversible isobaric heating process in a closed system Choose one answer in each line W12 = p(V1 – vz) O W12 = 0 W12 = p(v2 – v,) E W12 = v(P1 – P2) 0 912 = U2 – U1 912 = h2 – h,0 912 = T(S2 – S1) O 912 = 0 S2 - S1 < 0 S2 - S1 = 0 O S2 - S1 > 0

In a reversible isobaric heating process in a closed system Choose one answer in each line W12 = p(V1 – vz) O W12 = 0 W12 = p(v2 – v,) E W12 = v(P1 – P2) 0 912 = U2 – U1 912 = h2 – h,0 912 = T(S2 – S1) O 912 = 0 S2 - S1 < 0 S2 - S1 = 0 O S2 - S1 > 0

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

In a reversible isobaric heating process in a closed system
Choose one answer in each line
W12 = p(v1 – vz) O
W12 = 0
W12 = p(v2 – v1)
W12 = v(P1 – P2) O
912 = U2 – U1
912 = h2 – h0
912 = T(S2 - S1) O
%3D
912 = 0
S2 - S1 = 0 O
S2- S1 > 0
0> Is – 2s

In a reversible isobaric cooling process in an open system:
Choose one answer in each line
W12 = p(v1 – v2) O
W12 = 0
W12 = p(v2 – v,) O
%3D
W12 = v(p1 – P2) O
%3D
912 = u2 – U1
912 = h2 – h,O
- h,0
912 = T(S2 – s1) O
%3D
912 =0
%3D
S2 - S1 <0
S2 - S1 = 0 O
S2 - S1 > 0

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