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a thermodynamic system is taken from the...

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) = 20` cal, what is for the process ib?

Text Solution

Verified by Experts

`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`
`Delta U = 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 = Delta U + W = 50.4 + 47 = 97.4 J`
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