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question

An ideal gas mixture consists of 1 mole of oxygen and moles of nitrogen at 90°F and 13.5psia. (a) If each constituent gas is assigned a datum-state value of zero entropy at 14.7psiaand 0°F calculate the entropy of the mixture. (b) Determine the change in internal energy and entropy if the mixture is cooled to 40°F at constant volume. Assume that the heat capacities are constant for the temperature range of interest and have the following valves: C_(p)oxygen = 7.01C_(v)oxygen = 5.02 C_(p)nitrogen = 6.96C_(v)nitrogen = 4.97

category

engineering

options

  1. Entropy of the mixture = 7.04 Btu/°R, Change in internal energy = -748 Btu, Change in entropy = -1.526 Btu/°R
  2. Entropy of the mixture = 9.04 Btu/°R, Change in internal energy = -848 Btu, Change in entropy = -1.626 Btu/°R
  3. Entropy of the mixture = 7.54 Btu/°R, Change in internal energy = -848 Btu, Change in entropy = -1.226 Btu/°R
  4. Entropy of the mixture = 8.04 Btu/°R, Change in internal energy = -748 Btu, Change in entropy = -1.426 Btu/°R
  5. Entropy of the mixture = 8.54 Btu/°R, Change in internal energy = -848 Btu, Change in entropy = -1.526 Btu/°R
  6. Entropy of the mixture = 6.54 Btu/°R, Change in internal energy = -648 Btu, Change in entropy = -1.326 Btu/°R
  7. Entropy of the mixture = 10.04 Btu/°R, Change in internal energy = -948 Btu, Change in entropy = -1.726 Btu/°R
  8. Entropy of the mixture = 9.54 Btu/°R, Change in internal energy = -648 Btu, Change in entropy = -1.626 Btu/°R
  9. Entropy of the mixture = 5.04 Btu/°R, Change in internal energy = -548 Btu, Change in entropy = -1.046 Btu/°R
  10. Entropy of the mixture = 6.04 Btu/°R, Change in internal energy = -648 Btu, Change in entropy = -1.246 Btu/°R

src

stemez-Thermodynamics

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