a thermodynamic system is taken from the state i to statef along the path iaf. It is found that the heat absorbed by the system is 40 cal and work done by the system is 16.19 cal. Along the path i bf, the heat absorbed is 35 cal. (a) Find the work done along the path i bf. (b) If W = - 10 cal for the curved return path fi, what is for this path ?(c) Take U_(b) = 20 cal, what is U _(f)? (d) If U _(b) = 20cal, what is for the process ib?

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1.	a thermodynamic system is taken from the state i to statef along the path iaf. It is found that the heat absorbed by the system is 40 cal and work done by the system is 16.19 cal. Along the path i bf, the heat absorbed is 35 cal. (a) Find the work done along the path i bf. (b) If W = - 10 cal for the curved return path fi, what is for this path ?(c) Take U_(b) = 20 cal, what is U _(f)? (d) If U _(b) = 20cal, what is for the process ib?
Answer» Solution :`1 cal = 4.2` Joule First law of thermodynamilcs is `Delta U = Q - W = U _(f ) - U _(i)` `Q = ( U _(f) - U _(f)) + W` ` Q = 40 cal = 40 xx 4.2` `= 168 J` `W = 16.19 cal` `= 16. 19 xx 4.2 = 68 J` `U _(f) - U _(i) = Q - W` `= 168 - 68 = 100 J` The CHANGE in internal ENERGY between i and f is 100 J, and is independent of the path followed, because U is a point function and not a path function (a) Work done along the path i b f ` W = Q - Delta U` `Q = 35 xx 4.2=147 - 100 = 47 J` (b) For the path `fi, Q = ? W =- 10 cal =- 42 J` `Q = W + ( U_(i)U _(f)) = W - (U _(f) - U _(i)) =0 42 -100 = 0 142 J` (c ) `U _(i) = 8 cal = 8 xx 4.2 = 33.6 J U _(f) = ?` `U _(f) - U _(i) = 100 J` (d) `U _(b) = 20 cal = 20 xx 33.6 = 133. 6 J` `U _(i) = 8 cal = 8 xx 4.2 = 33. 6 J` `DeltaU = U _(b) - U _(i) = 84 - 33. 6 = 50.4 J` The work done for the path ib is same as that for i b f so `W = 47 J `from (a) `Q = DeltaU + W = 50.4 + 47 = 97.4 J`