One mole helium is enclosed in a cylinder with a moveable piston. By placing the

Question

One mole helium is enclosed in a cylinder with a moveable piston. By placing the cylinder in contact with various reservoirs and also insulating it at the proper times, the helium preforms the cycle shown m the figure below, process AB is isothermal, BC is adiabatic, CD is isobaric, and DA is isochoric. a. Compute the internal energy change, heat transfer, and work performed for each portion of the cycle. b. Determine total amount of each of these quantities for entire cycle and efficiency of the system. Assume helium to be an ideal gas and C_a = 20 B J/mole.K and C-y = 12.5 J/mole.K

Explanation / Answer

The process AB is constant temperature process and hence change in internal energy = 0

Temperature during isothermal process is From gas law equation, PV= nRT n= no of moles= 1, PA= 8*104 Pa =80Kpa (1000Pa= 1Kpa)= 800/101.3 atm (1013. Kpa= 1atm)=7.9 atm, VA= 2*10-3m3

= 2*10-3*1000L =2 L, R=0.0821 L.atm/mole.K TA= TB= PV/nR=7.9*2/0.0821 = 192 K

Work done during isothermal process , W = -nRT ln( VB/VA) =- 1*8.314 Joules* 192* ln(5/2) =1463 joules

{ VB =volume at the end of constant temperature process at point B and VA= volume at the begining of the process at point A) . Since because there is expansion of gas ,system does work on the surroundings and work done is –ve.

From 1st law of thermodynamics, delaU= Q+W, 0= Q+W and Q=-W= 1463 joules

Pressure at point B can be calculated as PAVA= PBVB, PB= PA*VA/VB= 8*104*2*10-3/ (5*10-3)= 3.2*104

b) path BC is adiabatic process and for adiabatic process, Q=0 , W= R*(TC-TB)/ (Y-1)

Y= CP/CV= 20.8/12.5=1.66

TC= temperature at point C , TB= temperature at point A= 192K

TC/TB= (PC/PB) Y-1 = (3/3.2)(1.66-1) , TC= 184 K

Work done = 8.314*(184-192)/(1.66-1)= -101 Joules

c) path CD is isobaric ( constant pressure process). For constant pressure process, W=-n*R*(TD-TC),deltaU= change in internal energy= n*CV*(TD-TC) and Q= n*CP*(TD-TC)

for isobaric process, VC/TC= VD/TD, TD= VD*TC/VC= 184*2/5= 74 K

hence work done = -1*8.314*( 74-184) =915 joules. Change in internal energy =1*12.5*(74-184)=-1375 and Q= deltaU-W= -1375-915=-2290 joules

d) for isochoric process ( constant volume process), work done =0 , Q= n*Cv*(TA-TD)=1*12.5*(192-74)= 1475 joules and deltaU=0 and Q= -W= -1475 joules

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