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Expand 1.00 molof a monatomic gas initially at 5.00kPaand 600 Kfrom initial volumeVi=1.00m3to final volumeVf=2.00m3. At any instant during the expansion, the pressure pand volume Vof the gas are related byp=5.00exp[Vi-VIa], with pin kilopascals,Vi and Vin cubic meters, anda=1.00m3. (a) What is the final pressure and (b) what is the final temperature of the gas? (c) How much work is done by the gas during the expansion? (d) What isSfor the expansion? (Hint: Use two simple reversible processes to findS.)

Short Answer

Expert verified

a) The final pressure of the gas is 1.84 kPa

b) The final temperature of the gas is 441 K

c) Work is done by the gas during expansion is 3.16 kJ

d) Entropy change for the expansion is 1.94 J/K

Step by step solution

01

The given data

Number of moles of monatomic gas, n = 1 mol

Initial pressure,pi=5kPa

Initial temperature,Ti=600K

Initial volume,Vi=1.00m3

Final volume,Vf=2.00m3

Value of a,a=1.00m3

Use two reversible processes to findS

02

Understanding the concept of thermodynamic equations 

We use the given condition to calculate the final pressure. Using the ratio form of the ideal gas equation, we can find the final temperature. We can use the formula of work done and entropy change of ideal gas to find the corresponding quantities.

Formulae:

The ideal gas equation,

pV=nRT (1)

The work done for an isobaric process,

W=ifpdV (2)

The entropy change of the gas,

S=nRInVfVi+nCvInTfTi (3)

03

(a) Calculation of the final pressure of the gas

The final pressure of the gas is given as:

pf=5kPaexpVf-Vi/a=5kPaexp2.00m3-1.00m3/1.00m3=5kPae-1=5kPa×0.3678=1.84kPa

Hence, the value of the final pressure is 1.84 kPa

04

(b) Calculation of the final temperature of the gas

Taking the ratio of two equations for the initial and final condition of the gas using equation (1), we get that the final temperature is given as:

pfVfpiVi=nRTfnRTiTfTi=pfVfpiViTf=pfVfpiViTiTf=600K1.84kPa2.00m35kPa1.00m3Tf=441K

Hence, the value of the final temperature is 441 K

05

(c) Calculation of work is done by the gas

Substituting the given value of pressure in equation (2), we get that the work done by the gas is given as:

W=5kPavivfeVi-V/adV=5kPaeVi/avivfe-V/adV=5kPaeVi/a-aeVaVi=1.00m3Vf=2.00m3=5kPae1.00m31.00m3-1.00m3e-2.00m3/1.00m3-e-1.00m3/1.00m3=5kPae1-1.00m3e-2-e-1

W=5kPae11.00m3e-1-e-2=5kPa2.71821.00m30.3678-0.1353=5kPa2.71821.00m30.2324=3.16kJ

Hence, the value of the work done by the gas is 3.16 kJ

06

(d) Calculation of the entropy change for the expansion

Entropy change for an ideal monatomic gas using equation (3) and the given values are given as follows:

S=nRIn2.00m31.00m3+n32RIn2e-1=piViTiIn2+32In2+32Ine-1=5000Pa1.00m3600KIn2+32In2+32Ine-1=5000Pa1.00m3600K0.2328=1.94J/K

Hence, the value of the entropy change of the gas is 1.94 J/k

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