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 the complete cycle. Express your answer in terms of the gas constant R.

Answer

A, B, C

Explanation

(a) Since AB is a straight line in V-T graph
`:. V/T=Constant(Isobaric process)`
`:. (V_A)/(T_A)=(V_B)/(T_B)`
`T_B=(V_B)/(V_A)xxT_A=2xx300=600K [:' (V_B)/(V_A)=2]`
(b) (i) A to B is a isobaric process
`:. Q=nC_pDeltaT=2xx5/2Rxx300`
`=1500R [:' C_p=5/2R for monoatomic gas]`
(ii) B to C is an isothermal process.
Since the temperature is not changing
`:. Internal energy change dU=0`
`:. From first law of thermodynamics dQ=dW`
`:. Q=2.303xxnRTlog_(10)(V_f)/(V_i)`
`=2.30xx2xxRxx600xxlog_(10)2`
`=2763.6xxlog_(10)2xxR=831.8R`
(iii) C to D is a isochoric process
`:. dW=0`
`:. Q=nC_vDeltaT=n(3/2R)(T_A-T_B)`
`=2xx3/2Rxx(-300)=-900R`
Volume is constant as heat is released.
(iv) D to A is an isothermal process
`:. Q=2.303xxnRTlog_(10)(V_f)/(V_i)`
`=2.303xx2Rxx300xxlog_((V_f)/(V_i))=-831.8R`
Heat is released as Q is positive.
(c) Total work done
`=Q_(AtoB)+Q_(BtoC)+Q_(CtoD)+Q_(DtoA)`
`=(1500R+831.8R)-(900R+831.8R)=600R`