2. A cylinder with a tight-fitting piston has an initial volume of 1.00 m2. The

Question

2. A cylinder with a tight-fitting piston has an initial volume of 1.00 m2. The cylinder contains 10.8 moles of oxygen gas at 280.0K. Assume oxygen behaves as an ideal diatomic gas near room temperature. R 8.3 145 J/(mol-K) = 0.08206 Latn(mol-K) k-1.38 x 1023 J/K a) What is the initial pressure of the gas? b) What is the initial internal energy of the gas? c) Suppose the gas expands adiabatically to a volume of 2.00 m3 What is the pressure in the gas now? d) How much work is done on the gas during this adiabatic expansion? e) The gas is returned to its initial state through an isobaric compression and then an isovolumetric process. Draw a PV diagram of this cyclic process. ) What is the amount of heat Q that flows into the gas during one cycle? 010

Explanation / Answer

2. Vi = 1 m^3

n = 10.8 mol

oxygen

Ti = 280 K

R = 8.3145

k = 1.38*10^-23

ideal diatomic gas

a. Pi = nRTi/Vi = 25129.44 Pa

b. initial internal energy = nCvTi

now, Cp/Cv = gamma

Cp - Cv = R

hence

Cp = R + Cv

R + Cv = Cv*gamma

Cv = R/(1 - gamma)

gamma = 1.4

hence

U = -nRTi/(1 -gamma) = 62823.6 J

c. adiabatic expansion

Pf = ?

Vf = 2 m^3

hence

PiVi^1.4 = PfVf^1.4

Pf = 9522.277 Pa

d. adiabatic work = PiVi^(gamma)(Vf^1-gamma - Vi^(1-gamma))/(1 - gamma) = 15212.214356 J

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