The given figure shown a change of state `A` to state `C` by two paths `ABC` and `AC` for an ideal gas. Calculate the :

(a) Path along which work done is least.
(b) Internal energy at `C` if the internal energy of gas at `A` is `10 J` and amount of heat supplied to change its state to `C` through the path `AC` is `200 J`.
(c) Amount of heat supplied ot the gas to go from A to B, if internal energy of gas at state `B` is `10 J`.

Explanation

We know,
Path CA - Isothermal compression
Path AB - Isobaric expansion
Path BC - Isochoric change
Let `V_(1)` and `V_(f)` are initial volume and final volume at respective points.
For temperature `T_(1)` (For C):
`PV = nRT_(1)`
`2 xx 10 = 1 xx 0.0821 xx T_(1)`
`T_(1) = 243.60` K
For temperature `T_(2)` (For C and B)
`(P_(1)V_(1))/T_(1) = (P_(2)V_(2))/T_(2)`
`=(2 xx 10)/(T_(1)) = (20 xx 10)/T_(2)`
`therefore T_(2)/T_(1)=10 therefore T_(2) = 243.60 xx 10 = 24360` K
Path CA : `w=+2.303 n RT_(1) log V_(1)/V_(f) = 2.303 xx 1 xx 2 xx 243.6 log 10/1 = +1122.02` cal
`DeltaE =0` for isothermal compression, Also q=Q
Path AB `W =-P(V_(t)-V_(i))`
`=20 xx (10-1) = -180` litre atm
`=(-180 xx 2)/(0.0821) = -4384.9` cal
Path BC, `W =-P(V_(t)-V_(i))=0 (therefore V_(t)-V_(i)=0)` since volume is constant
For monoatomic gas heat charge at constant volume `=q_(v) = DeltaE`
Thus for path BC.
`q_(v) = C_(v) xx n xx DeltaT = DeltaE`
`therefore q_(v) = 3/2 R xx 1 xx (2436 - 243.6) = 3/2 xx 2 xx 1 xx 2192 A = 6577.2 cal`
Since process involves cooling
`q_(v) = DeltaE = -6577.2` cal
Also in path AB, the internal energy in state A and state C is same. Thus during path AB, an increase in internal energy equivalent of change in internal energy during path BC should take place. Thus, `DeltaE` for path AB = +6572.2 cal Now q for path `AB = DeltaE - W_(AB) = 6577.2 + 4384.9 = 10962.1` cal
Cycle `DeltaE =0`,
`q=-W = -[W_("path CA") + W_("path AB") + W_("path BC"))`
`=-[+1122.02 + (-4384.9)+0]`
`q=-W = +3262.88` cal