A) \[dH=TdS-Vdp\]
B) \[dH=SdT+Vdp\]
C) \[dH=-SdT-Vdp\]
D) \[dH=dE-pdV\]
Correct Answer: A
Solution :
Given, \[dE=TdS-pdV\] ... (i) \[H=E+pV\] ... (ii) Differentiating Eq. (ii) \[dH=dE+pdV+Vdp\] ... (iii) From Eq.,(i) and (iii), we get \[dH=TdS-Vdp\]You need to login to perform this action.
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