A) \[\frac{3}{2}{{N}_{a}}{{k}_{B}}\left( {{T}_{2}}-{{T}_{1}} \right)\]
B) \[\frac{3}{4}{{N}_{a}}{{k}_{B}}\left( {{T}_{2}}-{{T}_{1}} \right)\]
C) \[\frac{3}{4}{{N}_{a}}{{k}_{B}}\frac{{{T}_{2}}}{{{T}_{1}}}\]
D) \[\frac{3}{8}{{N}_{a}}{{k}_{B}}\left( {{T}_{2}}-{{T}_{1}} \right)\]
Correct Answer: D
Solution :
[d] From first law of thermodynamics \[\Delta Q=\Delta U+\Delta W=\frac{3}{2}.\frac{1}{4}R\left( {{T}_{2}}-{{T}_{1}} \right)+0\] \[=\frac{3}{8}{{N}_{a}}{{K}_{B}}({{T}_{2}}-{{T}_{1}})\,\,\left[ \because \,\,K\,\,=\frac{R}{N} \right]\]
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