 Hello everyone, Myself RC Biradar working as assistant professor in Walsh and Instruct Technology, Solapur. So, in this video we will see about how to calculate the work done in reversible adiabatic process. At the end of this session, students will be able to determine the work done in reversible adiabatic process. Let us see how to derive the formula for work done and before that we will going to prove the law of reversible process that is reversible adiabatic process is p raised to gamma equal to constant. So, now for infinitesimal process for infinitesimal process we know that dq is equal to dw plus du and we call it as equation number 1 and for adiabatic process we know that dq that is heat transfer is 0 and if you put this in the first equation we get dw and du is equal to 0 and also we know that the work done is equal to p into dv and change in internal energy is equal to mcv into dt and put these two variables in this above equation. So, we get p into dv plus mcv into dt is equal to 0 here if you replace this dt term we can use I have from ideal gas law we know that pv is equal to mrt and on differentiating the above equation. So, we get p into dv plus v into dp is equal to mr into dt. So, therefore, we get dt as p into dv plus v into dp divided by m into r. So, put this dt value in the above equation. So, in the second equation so, the second equation becomes p into dv as it is plus m into cv as it is and the dt term is p into dv plus v into dp divided by m into r. So, mm get cancelled here and on simplification again we get cv by r in bracket p into dv plus v into dp. So, we need to replace this term cv by r to replace that. So, we know that the mayor's equation is cp minus cv is equal to r and from this if I take cv common and we get here cv by cv and that ratio we know it as gamma and here 1 equal to r and from this cv by r is equal to 1 divided by gamma minus 1. So, put this in the above equation. So, in continuation with that so, I can write the equation as p into dv plus 1 divided by gamma minus 1 in bracket of p into dv plus v into dp. So, taking LCM of gamma minus 1 and we get we can simplify that as gamma minus 1 p into dv plus p into dv plus v into dp. So, open up this bracket we get p into dv gamma minus p into dv plus p into dv plus v into dp. So, this p dv and this p dv get subtracted and we remain with gamma into p dv plus v into dp equal to 0 and divide by p into v term on both the sides. We get this above equation as gamma into dv by v plus dp by p and this we write it as equation number 3 and on integrating equation 3 we get this as integration of 0 is equal to this as constant term integration of dv by v plus integration of dp by p and on integration we get this as a constant of integration called log c to the base e is equal to gamma remains constant and the integration of this is log v to the base e plus for this term log p to the base e. So, from the logarithmic rule I can multiply this p and v I can simplify this equation as p into v rest to gamma is equal to log c to the base e and cancelling log term from both the sides we remain with p v rest to gamma is equal to constant and this is the law of this is the law of reversible adiabatic process reversible adiabatic process this is hence proved now with this we can see how to find the work done in a reversible adiabatic process and in a reversible adiabatic process we know that a process path is like this between state 1 and state 2 and this is at a higher pressure p 1 and this is at a pressure p 2 and this is at v 1 and this is at v 2 pressure and this process path followed by p v rest to gamma equal to constant law and this is a reversible adiabatic process reversible adiabatic now I know that the equation for work done is integration of p into dv and we know that the reversible adiabatic law is p v rest to gamma is equal to constant and therefore p is equal to what c by v rest to gamma I can use and put this value of p in the first equation I get w as integration of c by v rest to gamma into dv and this term is constant so hence I can take this term outside I get integration of this as w so and I can write this as v rest to minus gamma into dv so the integration of this between state 1 to 2 will give you the work done for the reversible adiabatic process and it can be solved as c into the integration of this term is minus gamma plus 1 divided by minus gamma plus 1 between states 1 to 2 and I can change the states so this value changes to gamma minus 1 state between 2 to 1 and I know that c is p into v rest to gamma so replace that by p into v rest to gamma in bracket we have gamma minus gamma plus 1 by gamma minus 1 between states 1 to 2 and solving this equation so if I take this v rest to gamma in insert this bracket so I get v rest to gamma into v rest to minus gamma so into v rest to 1 divided by gamma minus 1 between states 2 to 1 so now simplifying we get p v divided by gamma minus 1 between states 1 to 2 and applying the limits we get p 1 v 1 minus p 2 v 2 between state 2 and 1 so divided by gamma minus 1 and this equation we derived as a work done equation for the reversible adiabatic process and since we know that p v is equal to m r t for ideal gas so this equation can be replaced as w equal to m r t 1 minus m r t 2 divided by gamma minus 1 and this can be simplified as m r into t 1 minus t 2 divided by gamma minus 1 this is a equation for work done and applying first law of thermodynamics first law of thermodynamics we know that q equal to w plus delta u and since q equal to 0 an adiabatic process this equation reduces to w equal to minus delta u and this can be equated equated to u 1 minus u 2 and this can be I can write it as m into c v so t 1 minus t 2 and this will give you a work done in terms of change in internal energy thank you