A) \[(T+2.4)K\]
B) \[(T-2.4)K\]
C) \[(T+4)K\]
D) \[(T-4)K\]
Correct Answer: D
Solution :
In an adiabatic process \[Q=0\] So, from the first law of thermodynamics \[W=-\Delta U=-n{{C}_{v}}\Delta \Tau =-n\left( \frac{R}{\gamma -1} \right)\] \[\Rightarrow \]\[W=\frac{nR}{\gamma -1}({{T}_{i}}-{{T}_{f}})\] (i) Hence \[W=6R\,J,n=1\,mol,\] \[R=8.31\,J/mol-K,\gamma =\frac{5}{3},{{T}_{i}}=TK\] Substituting given values in Eq. (i), we get \[\therefore \] \[6R=\frac{R}{5/3-1}(T-{{T}_{f}})\] \[\Rightarrow \]\[6R=\frac{3R}{2}(T-{{T}_{f}})\] \[\Rightarrow \]\[T-{{T}_{f}}=4\] \[\therefore \]\[{{T}_{f}}=(T-4)K\]
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