A Flexible Balloon A flexible balloon contains 0.320 mol of an unknown polyatomi

Question

A Flexible Balloon A flexible balloon contains 0.320 mol of an unknown polyatomic gas. Initially the balloon containing the gas has a volume of 7450 cm^3 and a temperature of 28.0 degree C. The gas first expands isobarically until the volume doubles. Then it expands adiabatically until the temperature returns to its initial value. Assume that the gas may be treated as an ideal gas with C_p = 33.26 J/mol middot K and gamma = 4/3. What is the total heat Q supplied to the gas in the process? What is the total change in the internal energy Delta U of the gas? What is the total work W done by the gas? What is the final volume V?

Explanation / Answer

For Isobaric process:

P = constant

V/T = constant

Volume double hence T also doubles. (deltaT = 2T - T = T)

Cv = Cp - R = 33.26 - 8.314 = 24.95 J/mol K

deltaU = n Cv deltaT = 0.320 x 24.95 x (273 + 28) = 2402.8 J

W = P deltaV = n R deltaT = 0.320 x 8.314 x (273 + 28) = 800.8 J

Q = deltaU + W = 3203.6 J

for Adiabatic Process:

Vi = 2 x 7450 x 10^-6 m^3 = 0.0149 m^3

PV = n R T

Pi = (0.320 x 8.314 x (273 + 28)) / (7450 x 10^-6) = 107490.5 Pa

Ti = 273 + 2(28) = 329 K

Tf = 273 + 28 = 301 K

for adiabatic process, Q = 0

W = Pi Vi^y [ Vf ^(1-y) - Vi^(1-y)] / (1 - y)

y = 4/3

and T V^(y-1) = constant

329 ( 2 x 7450 x 10^-6) ^ (4/3 - 1) = 301 ( V)^(4/3 - 1)

Vf = 0.0195 m^3

W = ( 107490.5 x (0.0149)^4/3 ) [ 0.0195^(1- 4/3) - 0.0149^(1-4/3) ] / (1 - 4/3)

W = 412.16 J

deltaU = - 412.16 J

A) Q = 3203.6 J

B) deltU = 2402.8 + (-412.16) = 1990.64 J

c) W = 800.8 + 412.16 = 1212.96 J

d) V = 0.0195 m^3

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