23. A piston/cylinder arrangement contains one mole of an ideal gas (the system)

Question

23. A piston/cylinder arrangement contains one mole of an ideal gas (the system) initially at 10.0 atm pressure and 300 K as shown in the accompanying illustration. Neglecting the mass of the piston, neglecting friction, and assuming iso- thermal conditions throughout, the pin restraining the piston is removed. For the resulting process: Air at 1.0 atm 0.750 m Weight (a) What is Q if the mass of the weight is zero? (b) What is Q if the mass of the weight is 100 kg? (c) What is mass of the weight is 254.35 kg? (d) What is Qst if the mass of the weight is 1017.4 kg? Restrining pin l mol 300 K 0.250 m 10.0 atm Data: g = 9.807 m s-2 (R)

Explanation / Answer

For Isothermal process:

work done = nRTlnV2/V1 or P1V1lnV2/V1 or P1V1lnP1/P2

Here we have P1 = 10 atm

1 atm = 101325 Pa

10 atm = 1013250 Pa

Area of cylinder is constant

Initial volume V1 = lenght x area of cross section = 0.25Am3

and using ideal gas equation

V1 =nRT/P = 1 x 8.314 x 300/(10 x 101325) = 0.00246158 m3

0.25A = 0.00246158 m3

A = 0.00984632 m2

V2 = 4V1 as can be seen from figure

a)

Mass is zero

W = 10 x 101325 Pa x 0.00246158 m3 ln(4V1/V1)

W = 3457.689 J ~ 3457.69 J

b)

If mass is 100 kg

force due to 100kg = mg = 100kg x 9.807 ms-2 = 980.7 kg m/s2 or N

Pressure due to this = Force / area = 980.7 N / 0.00984632 m2 = 99600.66 Pa

Pressure = 0.98298 atm

total final pressure = 1 atm + 0.98298 atm = 1.98298 atm

W = 10 x 101325 Pa x 0.00246158 m3 ln(10 atm /1.98298 atm)

W = 4035.567 J ~ 4035.57 J

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