A) \[(T+2.4)K\]
B) \[(T-2.4)K\]
C) \[(T+4)K\]
D) \[(T-4)K\]
Correct Answer: D
Solution :
Key Idea: In an adiabatic process, there is no heat transfer into or out of a system ie,\[Q=0\]. In an adiabatic process \[Q=0\] So, from first law of thermodynamics. \[W=-\Delta U\] \[=-n{{C}_{v}}\Delta T\] \[=-n\left( \frac{R}{\gamma -1} \right)({{T}_{f}}-{{T}_{i}})\] \[=\frac{nR}{\gamma -1}({{T}_{i}}-{{T}_{f}})\] ?.(i) Here: \[W=6R\text{ }J,\text{ }n=1\text{ }mol,\] \[R=8.31\text{ }J/mol-K,\text{ }\gamma =\frac{5}{3},\text{ }{{T}_{i}}=T\text{ }K\] 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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