Why two isentropic lines are not parallel in a that cycle?

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If u calculate the slope of isentropic process on a p-v diagram it comes out to be -gamma*p/v... Means the slope at each point is dependent on p and v values.. hence two isentropic processes cannot be parallel
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M.Tech (Thermal Engg)

Two is entropic lines can never intersect each other in a cycle
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Isentropic lines are assymptodes, which intersect at each other at infinity.
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HABITUAL

For a closed system undergoing an adiabatic reversible process entropy remains constant i.e. dS=dQ/T For adiabatic process dQ=0. hence dS=0, Hence two isentropic lines will be parallel but can't intersect each other.
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Tutor

@Ashwani:- dQ is a wrong expression. Heat transfer is a path function & is an inexact differential. So u can't express it like dQ. Rather, it should be written as del(Q). Also, if a process is isentropic, it doesn't always mean that it will be reversible adiabatic.....Mind it!!
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entropy=change in heat/ change in time . if i plot a graph as q in y axis and time as x axis for simliar values it will give parallel line for two isoentropic lines...please explain your question properly.
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because as per second law of thermodynamics it is not possible, if heat is coming in the body so while doing work a part of (heat)it reject in the surrounding (kelvin - Planks statement ) if adiabatic line intersect then heat is only coming in the body and due to intersection heat could not able to...
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because as per second law of thermodynamics it is not possible, if heat is coming in the body so while doing work a part of (heat)it reject in the surrounding (kelvin - Planks statement ) if adiabatic line intersect then heat is only coming in the body and due to intersection heat could not able to leave from the body so it violate 2nd law of Thermodynamics read less
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Chemical engineering subject tutor

For an adiabetic reversible process heat transfer is zero . since ds=DQ/DT we have Ds=O. Hence parallel but can't intersect.
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