We will assume that gas is ideal. So, the internal energy of air
U=Mm​CV​T
For air M=28.96⋅10−3kg/mol and CV​=2i​R=25​⋅8.31≈20.8mol⋅KJ​
U=Mm​CV​T=28.96⋅10−31​⋅20.8⋅293=210442J
The enthalpy of air
H=U+pV=210442+2â‹…106â‹…0.05=310442J
ΔU=Mm​CV​(T2​−T1​)→T2​=m⋅CV​ΔU⋅M​+T1​=
=1⋅20.8120000⋅28.96⋅10−3​+293=460K
T1​p1​V1​​=T2​p2​V2​​→V2​=T1​p2​p1​V1​T2​​=
V2​=T1​p2​p1​V1​T2​​=293⋅5⋅1062⋅106⋅0.05⋅460​=0.031m3
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