An aluminum (2.70 g/cm 3 ), 4.15-cm thick, 11.5-cm diameter cylindrical piston e

Question

An aluminum (2.70 g/cm3), 4.15-cm thick, 11.5-cm diameter cylindrical piston encloses a sample of a diatomic gas within a sealed vessel as shown. The piston, initially floating 34.3 cm above the bottom of the vessel, fits snugly against the smooth walls of the vessel, separating the gas from an empty volume above it. (a) If the gas is at 20.0 degrees celcius, how many particles does the sample comprise? (b) Suppose 4.00 J of energy are introduced into the system via heat. What is the piston’s displacement resulting from this heat exchange? (c) Calculate the change in the entropy of the gas caused by the heating.

Explanation / Answer

(A) V = pi r^2 L = pi (0.115/2)^2 (0.343)

V = 3.563 x 10^-3 m^3

T = 273 + 20 = 293 K

mass of piston = (2.70) (pi (11.5/2)^2 (4.15)) = 1163 g = 1.163 kg

P = Patm + (m g / A )

= (101325) + (1.163 X 9.8 / (pi (0.115/2)^2))

= 102423 Pa

Applying PV = n R T

(102423)(3.563 x 10^-3) = n (8.314) (293)

n = 0.15 mol

particles = n NA = 9.02 x 10^22 particles

(b) This will be a isobaric process.

Q = n Cp deltat

for diatomic gas: Cp = 7 R / 2

4 = (0.15)(7 x 8.314 / 2) (deltaT)

deltaT = 0.916

W = n R deltaT = 0.15 x 8.314 x 0.916 = 1.14 J

(c) for isobaric process,

deltS = Cp ln(T2 / T1)

= (7 x 8.314 / 2) ln(293.916 / 293)

= 0.091 J/K
  

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