Question #257942

Assume that a 5-lb ideal gas with š‘… = 38.7 š‘“š‘”.š‘™š‘ š‘™š‘.Ā°š‘… and š‘˜ = 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.


Expert's answer

(a)

m=5ā€…ā€ŠlbR=38.7ā€…ā€Šftā‹…lb/lbmā‹…Rk=1.668Q=300ā€…ā€ŠBTU=233451lbfā‹…ftT1=80Q=mcp(T2āˆ’T1)233451=5ƗkRkāˆ’1(T2āˆ’T1)233451=5Ɨ1.668Ɨ38.70.668Ɨ(T2āˆ’80)T2=563.165ā€…ā€ŠĀ°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)

Ī”U=mcv(T2āˆ’T1)=mRkāˆ’1(T2āˆ’T1)=5Ɨ38.70.668Ɨ(563.165āˆ’80)=139958.6331ā€…ā€Šlbfā‹…ftĪ”U = 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)

Ī”H=Q=233451ā€…ā€Šlbfā‹…ftĪ”S=mcpln(T2T1)=5Ɨ1.6680.668Ɨ38.7Ɨln(563.16580)=942.9295Ī”H = Q = 233451 \;lbf \cdot ft \\ Ī”S = 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=Qāˆ’Ī”U=233451āˆ’139958.633=93492.3669ā€…ā€Šlbfā‹…ftW = Q -Ī”U \\ = 233451 -139958.633 \\ = 93492.3669 \; lbf \cdot ft


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