A monoatomic ideal gas of two moles is taken through a cyclic process starting from `A` as shown in figure. The volume ratios are `(V_(B))/(V_(A))=2` and `(V_(D))/(V_(A))=4`. If the temperature `T_(A)` at `A` is `27^(@)C`, calculate:

(a) The temperature of the gas at point `B`.
(b) Heat absorbed or released by the gas in each process.
(c) The total work done by the gas during complete cycle.

Answer

4988

Explanation

Given
Number of moles, n = 2
`C_V = 3/2 R " and " C_P = 5/2 R " (monoatomic)"`
`T_A = 27^@ C = 300 K`
Let `V_A =V_0` then `V_B = 2V_0` and `V_D = V_C = 4V_0`
Process `A to B`
`:." "Q_("AB") = nC_p Dt nC_p (T_B -T_A)`
`=(2)(5/2 R) (600 - 300)`
`Q_("AB") = 1500 " (absorbed)"`
Process `B to C`
` T = "constant ":."dU = 0"`
`:." "Q_("AB") = W_("BC") = nRT_B "In" (V_C/V_B)`
`=(2)(R) (600)"In"((4V_0)/(2V_0))`
`= (1200 R)In (2) = (1200R)(0.693)`
or `Q_("BC") ~~ 831.6 R ` (absorbed)
Process `C to DV=` constant
`:." "Q_(CD) = nC_V dT = nC_v (T_D - T_C)`
`= n(3/2 R)(T_A -T_B)(T_D =T_A"and"T_C =T_C)`
`=(2)(3/2 R)(300 - 600)`
`Q_("CD") = -900 R` (released)
Process `D to A` T = constant ` rArr Delta U = 0" ":." "Q_("DA")= W_("DA") = nRT_0 "In" (V_A/V_D)`
`Q_("DA") ~~-931.6R` (released)