One-quarter lbmol of oxygen gas (O2) undergoes a process from p 1 = 20 l b f / i n 2 , T 1 = 500 ∘ R to p 2 = 150 l b f / i n 2 . For the process W = -500 Btu and Q = -152.5 Btu. Assume the oxygen behaves as an ideal gas.

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One-quarter lbmol of oxygen gas (O2) undergoes a process from       p    1    =  20     l      b    f        /    i      n    2    ,         T    1    =      500    ∘    R   to       p    2    =  150  l      b    f        /    i      n    2  . For the process W = -500 Btu and Q = -152.5 Btu. Assume the oxygen behaves as an ideal gas.

devilvunga · Accepted Answer

Given data: Initial Pressure       P    1    =  20     l  b      /    i      n    2  Initial Temperature       T    1    =      500    ∘    RFinal pressure        P    2    =  150     l  b      /    i      n    2  Work   W  =  −  500     B  t  uHeat   W  =  −  152.2     B  t  u  n  =      1    4    l  b     m  o  lSpecific heats for oxygen       c    v    =  0.155     B  t  u      /    l  b     R  ,         c    p    =  0.217     B  t  u      /    l  b     R    R  =      C    p    −      C    v      R  =  0.062     B  t  u      /    l  b     RWe know that       1    4    =      m    32    m  =  8     l  bThe change in entropy is given by   △  S  =  m  [      C    p    ln  ⁡  (            T      2              T      1        −  R  ln  ⁡  (            P      2              P      1        )  )  ]    =  8  [  0.217  ⋅  ln  ⁡  (      780.242    500    )  −  0.062  ln  ⁡  (      150    20    )  ]    =  −  0.226878     B  t  u      /    R

One-quarter lbmol of oxygen gas (O_2) undergoes a process from p_1=20(lb_f)/(i n^2), T_1=500^circ R to p_1=150(lb_f)/(i n^2). For the process W = -500 Btu and Q = -152.5 Btu. Assume the oxygen behaves as an ideal gas.

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