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.

Jazmyn Pugh 2022-09-25 Answered
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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Answers (1)

devilvunga
Answered 2022-09-26 Author has 14 answers
Given data: Initial Pressure P 1 = 20   l b / i n 2
Initial Temperature T 1 = 500 R
Final pressure P 2 = 150   l b / i n 2
Work W = 500   B t u
Heat W = 152.2   B t u
n = 1 4 l b   m o l
Specific 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   R
We know that
1 4 = m 32
m = 8   l b
The 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
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