Answer to Question #257942 in Mechanics | Relativity for giel

Question #257942

Assume that a 5-lb ideal gas with 𝑅 = 38.7 𝑓𝑑.𝑙𝑏 𝑙𝑏.°𝑅 and π‘˜ = 1.668 have 300 Btu of heat added during reversible constant pressure change of state. The initial temperature is 80℉. Determine (a) the final temperature; (b) the change of enthalpy, internal energy, and entropy; and (c) the work.


1
Expert's answer
2021-10-28T08:51:00-0400

(a)

"m=5 \\;lb \\\\\n\nR = 38.7 \\;ft \\cdot lb\/lbm \\cdot R \\\\\n\nk = 1.668 \\\\\n\nQ = 300 \\;BTU = 233451 lbf \\cdot ft \\\\\n\nT_1 = 80 \\\\\n\nQ = mc_p(T_2-T_1) \\\\\n\n233451 = 5 \\times \\frac{kR}{k-1}(T_2-T_1) \\\\\n\n233451 = 5 \\times \\frac{1.668 \\times 38.7}{0.668} \\times (T_2-80) \\\\\n\nT_2 = 563.165 \\; \u00b0F"

(b)

"\u0394U = mc_v(T_2-T_1) \\\\\n\n= \\frac{mR}{k-1}(T_2-T_1) \\\\\n\n= 5 \\times \\frac{38.7}{0.668} \\times (563.165-80) \\\\\n\n= 139958.6331 \\; lbf \\cdot ft"

(c)

"\u0394H = Q = 233451 \\;lbf \\cdot ft \\\\\n\n\u0394S = mc_pln(\\frac{T_2}{T_1}) \\\\\n\n= 5 \\times \\frac{1.668}{0.668} \\times 38.7 \\times ln(\\frac{563.165}{80}) \\\\\n\n= 942.9295"

(d)

"W = Q -\u0394U \\\\\n\n= 233451 -139958.633 \\\\\n\n= 93492.3669 \\; lbf \\cdot ft"


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