A) \[\Delta U=150\,RJ\,mo{{l}^{-1}},\] \[\Delta V=20\left[ 1-{{\left( \frac{3}{2} \right)}^{1.5}} \right]L\]
B) \[\Delta U=0,\,\,\Delta V=-20{{\left( \frac{3}{2} \right)}^{1.5}}L\]
C) \[\Delta U=-150RJ\,mo{{l}^{-1}},\,\,\Delta V=-20\,{{\left( \frac{3}{2} \right)}^{1.5}}L\]
D) \[\Delta U=0,\,\Delta V=20\left[ 1-{{\left( \frac{3}{2} \right)}^{1.5}} \right]L\]
Correct Answer: A
Solution :
[a] Adiabatic reversible compression: \[\Delta T=({{T}_{1}}-{{T}_{2}})=100K\] \[\Delta U={{C}_{v,m}}({{T}_{1}}-{{T}_{2}})=\frac{3}{2}R(100K)\] \[=150\,R\,J\,mo{{l}^{-1}}\] \[\Delta V={{V}_{1}}\left[ 1-{{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{Cv,m/R}} \right]\] \[=20L\left[ 1-{{\left( \frac{300}{200} \right)}^{3/2}} \right]\] \[=-16.74L\]You need to login to perform this action.
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