Solution.
m=1kg;
V1=0.05m3;
p1=2⋅106Pa;
ΔU=120⋅103J;
p2=5⋅106Pa;
M=28.96⋅10−3kg/mol;
T1=293K;
U=MmCVT;
CV=2iR=25⋅8.31=20.8J/molK;
U=28.96⋅10−31⋅20.8⋅293=210442J;
H=U+pV;
H=210442+2⋅106⋅0.05=310442J;
ΔU=MmCV(T2−T1)⟹T2=mCVΔUM+T1;
T2=1⋅20.8120⋅103⋅28.96⋅10−3+293=460K;
T1P1⋅V1=T2P2⋅V2⟹V2=T1P2P1V1T2;
V2=293⋅5⋅1062⋅106⋅0.05⋅460=0.031m3;
Answer: U=210442J;
H=310442J;
V2=0.031m3.
Comments
Leave a comment