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Time management is very much important in IIT JAM. The eduncle test series for IIT JAM Mathematical Statistics helped me a lot in this portion. I am very thankful to the test series I bought from eduncle.
Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
- IIT JAM
- Physics (PH)
Thermodynamics
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Abhishek singh Best Answer
Entropy of the gas remains unchanged (it is sometimes written as isentropic process) which clearly means dS =0, which means dQ =0 (as dQ = T*dS). Hence isentropic process is adiabetic process (along with an extra piece of information about the reversible nature of process). I will suggest you a time saving trick here, which is also known as smart work(as it saves your time in exam). After reaching at equation 1 (of my solution). you don't need to check all options. look at option A and D, in these options T and V are changing in such a way that their product remains constant, which is not suitable for our equation 1. so you can directly check option B and C. out of which option B holds the equation 1. So it is the correct answer. Best regards
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Shanu arora
Option B is currect.As dS=0 thus the process here is adiabatic. dQ=TdS .so using pv^ (gamma) = const , gives the solution .I have checked all the solution ,thinking it as a MSQ, feel freefor any doubts.
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Ruby negi![best-answer]()
This is a isoentropic process or you can say adibatic process also(dS=0 means dQ=0).. check all the options by putting them in equation of adibatic process, u will get ur ans...