A container contains 2.000 mol of mono-atomic gas at a pressure of 2.600atm and

Question

A container contains 2.000 mol of mono-atomic gas at a pressure of 2.600atm and occupies a volume of 20.00L at point A. (1) The gas is heated at constant pressure until volume becomes 40.00L at point B. the gas is cooled at constant volume until pressure decreases to 1.300atm at point C. the gas is compressed isothermally back to point A. 1.00 atm = 10^5 Pa; 1.00L = 10^'3 m^3. R = 8.314 J/mol.k What is temperature at points A, B and C? How much work is done on the system in each process? How thermal energy is added during the entire cycle.

Explanation / Answer

a) By ideal gas law,

Temperature at A = PV/nR = 2.6*10^5*20*10^-3/(2*8.314) = 312.7 K

Temperature at B = PV/nR = 2.6*10^5*40*10^-3/(2*8.314) = 625.4 K

Temperature at B = PV/nR = 1.3*10^5*40*10^-3/(2*8.314) = 312.7 K

B) Wab = P delta V = 2.6*10^5*[(40-20)*10^-3] = 5200 J

Wbc = P delta V = 0

Wca = nRT ln (Va/Vc) = 2*8.314*312.7*ln(0.5) = -3604 J

C) thermal energy added = Wab + Wbc+Wca = 5200-3604 =1596 J

  

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