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Question
A diatomic ideal gas undergoes a thermodynamic change according to the P−V diagram shown in the figure. The total heat given to the gas is nearly (use ln2 = 0.7) :
Options
A.2.5 P0V0
B.1.4 P0V0
C.3.9 P0V0
D.1.1 P0V0
Solution
QAB = ᐃ UAB + WAB
WAB = 0
ᐃ UAB =
nR ᐃ T ⇒ 
(ᐃ PV)
ᐃ UAB = (ᐃ PV)
QAB = 2.5 P0 V0
Process BC
QBC = ᐃ UBC + WBC
QBC = 0 + 2P0 V0 ln 2 = 1.4 P0 V0
Qnet = QAB + QBC = 3.9 P0 V0
WAB = 0
ᐃ UAB =
nR ᐃ T ⇒ (ᐃ PV)ᐃ UAB = (ᐃ PV)
QAB = 2.5 P0 V0
Process BC
QBC = ᐃ UBC + WBC
QBC = 0 + 2P0 V0 ln 2 = 1.4 P0 V0
Qnet = QAB + QBC = 3.9 P0 V0
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