A monatomic ideal gas of two moles is taken through a cyclic process starting from A as shown in the 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 the complete cycle. Express your answer in terms of the gas constant R.

Explanation

Taking `V_A=V_0, V_B =2V_0 , V_D =4V_0`
Process `A to B` : Isobaric process as
V/T = constant ( `:.` VT graph is straight line)
`[(V_0)/((273 + 27 ))] =((2V_0))/(T_B)`
(or)` T_3 = 600 K = 327^@ C`
` Q_1 = nC_p Delta T = 2 `(5/2) R (600 - 300) = 1500 R
Heat is absorbed in this process as `Q_1` is positive .
Process `B to C `(Isothermal ) [ `:.` temperature is constant]
` Q_2 =W_2=nRT_B In (V_2 //V_1)`
`=2 xx R xx 600 xx log ((4V_0)/(2V_0)) = 1200 R log 2 = 831.78R`
Heat is absorbed in this process as Q, is positive. Process `C to D` (Isochoric) (`:.` Volume is constant)
`Q_3 = nC_v Delta T = 2 xs (3//2R) xx (300 - 600) = -900 R`
Heat is released in this process as `Q_3` is negative.
Process `D to A` (Isothermal)
` Q_4 =W_4=nRT_A In (V_2//V_1) = 2 xx R xx 300 xx log (V_0 //4V_0) = -831.78 R`
Total work done :` W=Q_1 +Q_2 +Q_3 + Q_4 = 1500 R+ 831.78 R-900 R - 831.78 R=+600 R.`