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An ideal gas after going through a series of four thermodynamic states in order, reaches the initial state again (cyclic process). The amounts of heat, Q and work, W involved in these states are Q, =6000J, 42 =-5500J , Q3 = -3000 J, Q, =3500 W, =2500J, W2 =-1000J, Ws =-1200J, IN, =x J The ratio of net work done by the gas to the total heat absorbed by the gas is r. Calculate the value of x and n.

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Total heat in this process=6000-5500-3000+3500 =1000J Total work in this process=w1+w2+w3+w4 =2500-1000-1200+x W=300+x Total heat=Total work Q=W ...
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Total heat  in this process=6000-5500-3000+3500

                                                =1000J

Total work in this process=w1+w2+w3+w4

                                                =2500-1000-1200+x

                                            W=300+x

Total heat=Total work

               Q=W

             1000=300+x

So.    x=700

n=net work done/total heat absorbed

n=1000×100/(6000+3500)=10.5%

 

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