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.

A monatomic ideal gas of two moles is taken through a cyclic process s

1.	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.
Answer» Solution : Taking `V_A=V_0, V_B =2V_0 , V_D =4V_0` Process `Ato 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_pDelta 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_BIn (V_2 //V_1)` `=2 xx R xx600 xx log ((4V_0)/(2V_0))= 1200 R log 2 = 831.78R` Heat is absorbed in this process as Q, is positive. Process `Cto 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 `Dto 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.`