One mole helium is enclosed in a cylinder with a moveable piston. By placing the
ID: 1509324 • Letter: O
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 in the figure below. Process AB is isothermal. BC is adiabatic. CD is isobaric. and DA is isochoric. Compute the internal energy change, heat transferred, and work performed for each portion of the cycle and total amount of each of these quantities for entire cycle. Assume helium to be an ideal gas and Cp=20.8 J/mole.K and Cv=12.5 J/mol.K.Explanation / Answer
Hi,
The way to calculate the work done by the gas or the job done over the gas is:
W = P dV
In this problem I am going to consider that:
- The work done by the gas is positive.
- The heat transferred to the gas is positive.
The way to calculate the internal energy change is:
U = Q - W
Process BC (adiabatic process)
In this case the ideal gas is doing work over the piston but it is nor receiving heat, therefore the temperature of the gas decreases.
The way to calculte the work is:
PVy = constant = C ::::::::: PBVBy = PCVCy = C
y = Cp/Cv = 20.8/12.5 = 1.664
C = (3*104Pa)(5*10-3 m3)1.664 = 4.45
W = C dV/Vy ::::::::: W = [C/(1-y)] [VC1-y - VB1-y] = (PCVC - PBVB)/(1-y)
To find the work we need the value of VB, said value can be found (alongside with PB) if we solve the following system:
PV1.664 = 4.45 (adiabatic process)
PV = 159.63 (ideal gas)
PB = 3.5*104 Pa ; VB = 4.56*10-3m3
So: W = [ (3*104)(5*10-3) - (3.5*104)(4.56*10-3) ]/(1-1.664) = 14.46 J (positive as the work is done by the gas)
Q = 0 (the process does not allow it)
U = -W = -14.46 J (the gas has lost part of its energy due to the work done over the piston)
Process AB (isothermic process)
In this case the ideal gas is doing work over the piston mantaining the same temperature over the process. To achieve that, heat of equal magnitude to the work done by the gas must be transferred to the gas.
The way to calculate the work is:
P = nRT/V ; where n = 1 mol ; R = 8.314 Pa*m3 /(mol*K) and T = 19.2 K
W = (nRT) dV/V :::::: W = nRT Ln(VB/VA) = 8.314*19.2 Ln(4.56/2.00) = 131.56 J
W = 131.56 J (positive as the work is done by the gas)
Q = W = 131.56 J (positive as the heat is transferred to the gas)
U = 0 (as Q = W in this case)
Process CD (Isobaric process)
In this case the piston is doing work over the ideal gas. If the pressure remains constant, the volume and the temperature should change.
The way to calculate the work is:
W = P dV :::::::: W = P (VD - VC) :::::::: W = (3*104)(2.00 - 5.00)10-3 = -90 J (negative as the work is done over the gas)
The heat can be calculated as:
Q = nCp (TD - TC) ; where T = PV/RT ::::::: TD = 7.22 K and TC = 18.04 K
Q = (1) (20) (7.22 - 18.04) = -216.4 J (negative, as the heat is getting out of the gas)
U = -216.4 + 90 = -126.4 J (negative, as the system is losing more energy in form of heat than the oneis gaining through work).
Process DA (Isochoric process)
In this case the work will be equal to cero and if the volume remains constant that means that the temperature should change (so, there must be heat).
W = 0 (the process does not allow it)
Q = nCv (TA - TD ) = 12.5(19.2 - 7.22) = 149.75 J (positive as heat is going towards the gas)
U = Q = 149.75 J (positive as the system is gaining energy through heat).
Total result
W = 14.46 + 131.56 - 90 + 0 = 56.02 J (so the total result is net work made over the piston by the gas)
Q = 149.75 - 216.4 + 131.56 + 0 = 64.91 J (so the total result is heat tranferred towards the gas)
U = Q - W = 8.89 J (the system gains energy because receives more heat than the work it gives)
Note: the total heat is bigger than the total work, this is something that always happens with this kind of machines. It is impossible to transform completely heat into work.
I hope it helps.
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