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atjari Group TitleBest ResponseYou've already chosen the best response.0
dw:1331452540087:dw
 2 years ago

atjari Group TitleBest ResponseYou've already chosen the best response.0
pls see the attachment for the question
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.1
Efficiency is the ratio of the energy given to the system, to the net work done by the system. Now, can you tell me how will you find the net work done by the system from the graph?
 2 years ago

SidharthSahoo Group TitleBest ResponseYou've already chosen the best response.0
Hey Mani come to chat box
 2 years ago

atjari Group TitleBest ResponseYou've already chosen the best response.0
Net work=P2(V1V2)delta U
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.1
Well, actually the net work done by the system in a cyclic process is the area of the closed graph. Lets see how much work we have done to the system From 1 to 2, the work done is P2(V2V1) From 2 to 3, no work is done From 3 to 1, the work done is (P3V2P2V1)/(gamma 1) The work done by us is the sum of all the above. The work done by the gas is : W(1,2)+W(2,3)W(3,1) Find the ratios of work done by gas, to work done by us. That should be the efficiency.
 2 years ago

atjari Group TitleBest ResponseYou've already chosen the best response.0
Can u explain hw u got (P3V2P2V1)/(gamma 1) for 3to1
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.1
The work done by a gas in an adiabatic process is : \[(Pinitial \times VinitalPfinal \times Vfinal)/(\gamma1)\] You can also prove this. Now, is it ok?
 2 years ago

atjari Group TitleBest ResponseYou've already chosen the best response.0
Ha k. Thanx a lot.
 2 years ago

atjari Group TitleBest ResponseYou've already chosen the best response.0
dw:1331455348197:dw
 2 years ago

Mani_Jha Group TitleBest ResponseYou've already chosen the best response.1
Hey, let me calculate and see. I will post as soon as I do this
 2 years ago

atjari Group TitleBest ResponseYou've already chosen the best response.0
Ha k. Thanx.
 2 years ago
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