Consider the following four-process cycle that is carried out on a system of mon

Question

Consider the following four-process cycle that is carried out on a system of monatomic ideal gas, starting from state 1 in which the pressure is 80.0 kPa and the volume is 2.00 liters.

Process A is an isothermal process that triples the volume;
process B is a constant volume process that returns the system to a pressure of 80.0 kPa;
process C is an isothermal process that returns the system to a volume of 2.00 liters;
and process D is a constant volume process that returns the system to state 1.

You should sketch a P-V diagram for this cycle, to help you understand it.

Complete the table below. For each process, and the entire cycle, find the heat added to the gas, the work done by the gas, and the change in internal energy.

please explain/show work

Explanation / Answer

n = 1

PROCESS A - > B

The total work done is the sum of the work done in each step.
Work done on the gas is given by the integral
W = - p dV from initial to final volume

Since the Temp does not change in first step,
W = - nRT*ln(Vb/Va) = - 80000 Pa * 2*10^-3 m^3 * ln(6/2) = - 175.78 J

(b)
Assuming ideal gas behavior, internal energy depends solely on temperature (and not on pressure and volume):
U = nCvT
So the change of internal energy is:
U = nCvT

Because we return to initial temperature after second step, the change of internal energy is zero
U = 0

(c)
Change of internal energy equals work done On the gas plus heat transferred to the gas:
U = W + Q
=>
Q = U - W = 0J - (-175.78J) = 175.78 J

PROCESS B ->C

W = - p dV from initial to final volume

Since the the volume does not change in first step, no work is done on the gas:
W = 0

Pa = 80 kPa

At constant Temp

PaVa = PbVb

Pb = Pa*Va/Vb = 80*2/6 = 26.67 kPa

Ta = PaVa/nR = 19 K

Tb = PbVb/nR = 19 K

At constant volume

Tb/Pb = Tc/Pc

Pc = 80 kPa

Tc= TbPc/Pb = 19*80/26.67 = 57 K

So the change of internal energy is:
U = nCvT

For one mole of an ideal monatomic gas (Cv=3/2R, Cp=5/2R)

U = 3/2R * (57-19) = 474 J

Change of internal energy equals work done On the gas plus heat transferred to the gas:
U = W + Q
=>
Q = U - W = 474 - 0 = 474 J

PROCESS C -> D

Since the Temp does not change in this step,
Wa = - nRT*ln(Vd/Vc) = - 80000 Pa * 6*10^-3 m^3 * ln(2/6) = 527.33 J

(b)
Assuming ideal gas behavior, internal energy depends solely on temperature (and not on pressure and volume):
U = nCvT
So the change of internal energy is:
U = nCvT

Because we return to initial temperature after second step, the change of internal energy is zero
Ua = 0

(c)
Change of internal energy equals work done On the gas plus heat transferred to the gas:
U = W + Q
=>
Qa = U - W = 0J - (-527.33J) = - 527.33 J

PROCESS D -> A

W = - p dV from initial to final volume

Since the the volume does not change in first step, no work is done on the gas:
W = 0

Pa = 80 kPa

At constant volume

Td/Pd= Ta/Pa

Td= TaPd/Pa = 19*240/80 = 57 K

So the change of internal energy is:
U = nCvT

For one mole of an ideal monatomic gas (Cv=3/2R, Cp=5/2R)

U = 3/2R * (19-57) = - 474 J

Change of internal energy equals work done On the gas plus heat transferred to the gas:
U = W + Q
=>
Q = U - W = -474 - 0 = -474 J

OVERALL

Q = Qa+Qb+Qc+Qd = - 351.55 J

W = Wa+Wb+Wc+Wd = 351.55 J

U = Ua+Ub+Uc+Ud = 0

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