A) \[395.4{}^\circ C\]
B) \[144{}^\circ C\]
C) \[887.4{}^\circ C\]
D) \[18{}^\circ C\]
Correct Answer: A
Solution :
Initial temperature \[{{T}_{1}}=18{}^\circ C=291K\] Filial volume \[{{V}_{2}}=\frac{{{V}_{1}}}{8}\] (where\[{{V}_{1}}\]is the initial volume) Now, for adiabatic expression \[{{T}_{1}}{{V}_{1}}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}\] Hence, \[{{T}_{2}}={{T}_{1}}{{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma -1}}\] \[=291\times {{8}^{1.4-1}}\] \[=291\times 2.297\] \[=668.4\text{ }K\] \[=395.4{}^\circ C\]
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