A) \[Ve\left( \frac{3a}{2R} \right){{T}^{2}}=cons\tan t\]
B) \[Ve\left( \frac{-3a}{2R} \right){{T}^{2}}=cons\tan t\]
C) \[T{{V}^{2}}=cons\tan t\]
D) \[V{{T}^{2}}=cons\tan t\]
Correct Answer: B
Solution :
given \[C={{C}_{V}}+3a{{T}^{2}}\] |
From of thermodynamics |
\[Q=\Delta U+W\] |
\[or\,C\Delta T={{C}_{V}}\Delta T+P(\Delta V)\] |
Or \[C={{C}_{V}}+P\left( \frac{\Delta V}{\Delta T} \right);\] or \[C={{C}_{V}}+P\left( \frac{dV}{dT} \right)\] |
So\[P\left( \frac{dV}{dt} \right)=3a{{T}^{2}}\] |
Also \[P=\frac{RT}{V},\]\[So\frac{RT}{V}\left( \frac{dV}{dT} \right)=3a{{T}^{2}}\] |
Or \[\int{\frac{dV}{V}=\frac{3a}{R}\int{TdT;}}\] \[or\,\ln \,V=\frac{3a}{2R}{{T}^{2}}\] |
Or \[V={{e}^{3a{{T}^{2}}/2R}}\] |
Or \[V{{e}^{\frac{-3a}{2R}{{T}^{2}}}}=Cons\tan t.\] |
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