A) \[(T+2.4)\text{ }K\]
B) \[(T-2.4)\text{ }K\]
C) \[(T+4)\text{ }K\]
D) \[(T-4)\text{ }K\]
Correct Answer: D
Solution :
Key Idea: in an adiabatic process, there is no heat transfer into or out of a system i.e., \[Q=0\] |
In an adiabatic process \[Q=0\] |
So, from 1st law of thermodynamics. |
\[W=-\Delta U\] |
\[=-n{{C}_{V}}\Delta T\] |
\[=-\left( \frac{R}{\gamma -1} \right)({{T}_{f}}-{{T}_{i}})\] |
\[=\frac{nR}{\gamma -1}({{T}_{i}}-{{T}_{f}})\] |
Here: \[W=6R\,\,\,J,\,\,n=1\,\,\,mol,\] |
\[R=8.31\,J/mol-K,\,\,\gamma =\frac{5}{3},{{T}_{i}}=T\,K\] |
Substituting given values in Eq. (i), we get |
\[\therefore \] \[6R=\frac{R}{(5/3-1)}(T-{{T}_{f}})\] |
\[\Rightarrow \] \[6R=\frac{3R}{2}(T-{{T}_{f}})\] |
\[\Rightarrow \] \[T-{{T}_{f}}=4\] |
\[\therefore \] \[{{T}_{f}}=(T-4)K\] |
Note: Adiabatic expansions of mono, dia and polyatomic gases are shown below |
1\[\to \] monoatomic |
2\[\to \] diatomic |
3\[\to \] polyatomic |
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