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A Sample of an ideal gas is expanded 1 m^3 to 3 m^3 in a reversible process for which P=KV^2,with K=6 bar/m^6.work done by the gas is
Dear Rajat, first converting all the parameters in SI units, i.e. unit of constant, K = 6 bar/m^6 1 bar = 10^5 N/m^2, => K = 6X10^5 N/m^8 so we can find out work from the equation, Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best Rajat !!!Regards,Askiitians ExpertsRohith Gandhi P
Dear Rajat,
first converting all the parameters in SI units, i.e. unit of constant, K = 6 bar/m^6
1 bar = 10^5 N/m^2, => K = 6X10^5 N/m^8
so we can find out work from the equation,
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best Rajat !!!Regards,Askiitians ExpertsRohith Gandhi P
thanks rohith bhaiya.but p is not equal yo K/V^2.you had solved it that way.but anyways i get the answer .Thanks a lot!!!!!!!!!!!!11i wish u help me in near future too.
if u still want some answer then integrate KV^2dV limits fron 1 to 3
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