An ideal gas (2.0 moles) is carried round a cycle as shown. If the process `b rarr c` is isothermal and `C_(V) = 3 cal//mol//K`. Determine
a. work done,
b. change in internal energy,
c. heat supplied to the system during processes `a rarr b`, `b rarr c` and `c rarr a`.

Explanation

a. Work done , W
process `a rarr b` is clearly isobaric at P = 4 atm.
`W = P (V_(b) - V_(a)) = 4 (16.4 - 82) = 32.8 L atm`
`implies W_(a - b) = 32.8 - 101 = 3313 J`
Process `b rarr c` is isothermal (given)
`W = 2.303 nRT log (V_(c )//V_(b))`
`implies W = 2.303 (P_(b) V_(b)) log (V_(c )//V_(b))`
`W = 2.303 (4 xx 16.4 xx 101) log (8.2//16.4)`
`W_(b rarr c) = - 4592.5 J`
Process `c rarr a` is isochoric (volume is constant)
`W_(c rarr a) = 0 J`
b. change in internal energy `Delta U`
Process `c rarr a : Delta U = nC_(v) Delta T`
`Delta U = nC_(v) ((P_(b) V_(b))/(nR) - (P_(a) V_(a))/(nR))` (since pV = nRT)
`= (C_(v))/(R ) (P_(0) V_(0) - P_(a) V_(a)) = (3)/(0.0821) (4 xx 16.4 - 4 xx 8.2)`
`implies Delta U = 1200` cal
Process `b rarr c` : isothermal, `Delta T = 0`
`implies Delta U = 0`
Process `c rarr Delta U` for complete cyclic process is zero because it depends on initial and final states i.e., temperature difference only.
`(Delta U)_(ab) + (Delta U)_(bc) +(Delta U)_(ca) = 0`
`implies 1200 + 0 + (Delta U)_(ca) = 0`
`implies (Delta U)_(ca) = - 1200` col
Heat supplied : `Delta H`
Using the first law of thermodynamics:
`(Delta H)_(ab) = (Delta U)_(ab) + W_(ab)`
`= (1200 + 3313 // 4.18) cal = 1992.5 "cal"`
`implies (Delta Q)_(ab) = 1992.5 cal`
`implies (Delta H)_(bc) = (Delta U)_(bc) + W_(bc)`
`= (0 - 4592.5 // 4.18) cal = 1098.7 cal`
`implies (Delta H)_(bc) = 1098.7 cal`
`(Delta H)_(ca) = (Delta U)_(ca) + W_(ca) = - 1200 + 0 = - 1200 cal`
`implies Delta H = 1200 cal`