Question #257942

Assume that a 5-lb ideal gas with 𝑅 = 38.7 𝑓𝑡.𝑙𝑏 𝑙𝑏.°𝑅 and 𝑘 = 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  lbR=38.7  ftlb/lbmRk=1.668Q=300  BTU=233451lbfftT1=80Q=mcp(T2T1)233451=5×kRk1(T2T1)233451=5×1.668×38.70.668×(T280)T2=563.165  °Fm=5 \;lb \\ R = 38.7 \;ft \cdot lb/lbm \cdot R \\ k = 1.668 \\ Q = 300 \;BTU = 233451 lbf \cdot ft \\ T_1 = 80 \\ Q = mc_p(T_2-T_1) \\ 233451 = 5 \times \frac{kR}{k-1}(T_2-T_1) \\ 233451 = 5 \times \frac{1.668 \times 38.7}{0.668} \times (T_2-80) \\ T_2 = 563.165 \; °F

(b)

ΔU=mcv(T2T1)=mRk1(T2T1)=5×38.70.668×(563.16580)=139958.6331  lbfftΔU = mc_v(T_2-T_1) \\ = \frac{mR}{k-1}(T_2-T_1) \\ = 5 \times \frac{38.7}{0.668} \times (563.165-80) \\ = 139958.6331 \; lbf \cdot ft

(c)

ΔH=Q=233451  lbfftΔS=mcpln(T2T1)=5×1.6680.668×38.7×ln(563.16580)=942.9295ΔH = Q = 233451 \;lbf \cdot ft \\ ΔS = mc_pln(\frac{T_2}{T_1}) \\ = 5 \times \frac{1.668}{0.668} \times 38.7 \times ln(\frac{563.165}{80}) \\ = 942.9295

(d)

W=QΔU=233451139958.633=93492.3669  lbfftW = Q -ΔU \\ = 233451 -139958.633 \\ = 93492.3669 \; lbf \cdot ft


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