A) \[-{{142}^{o}}C\]
B) 300 K
C) \[{{327}^{o}}C\]
D) 420 K
Correct Answer: A
Solution :
Key Idea: When the gas is suddenly compressed heat does not find time to flow in or out. When a system undergoes a change under the condition that no exchange of heat takes place between the system and surroundings, then such a process is called adiabatic process. In this case gas is suddenly compressed, hence it is adiabatic process, the relation between temperature (T) and pressure (P) is \[\frac{{{T}^{\gamma }}}{{{p}^{\gamma -1}}}=\]constant where \[\gamma \] is ratio of specific heats. Given, \[{{T}_{1}}={{27}^{o}}C=27+273=300\,K,\,{{p}_{1}}=p\] \[{{p}_{2}}=\frac{p}{8},\,\gamma =\frac{5}{3}\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{p}_{1}}}{{{p}_{2}}} \right)}^{\frac{\gamma -1}{\gamma }}}\] \[\Rightarrow \] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{8}{1} \right)}^{\frac{\frac{5-1}{3}}{\frac{3}{5/3}}}}\] \[={{(8)}^{0.4}}\] = 2.297 \[\Rightarrow \] \[{{T}_{2}}=\frac{{{T}_{1}}}{2.297}=\frac{300}{2.297}\] \[\Rightarrow \] \[=130.6\,K\simeq 131\,K\] \[\Rightarrow \] \[{{T}_{2}}=131-273\] \[=-{{142}^{o}}C\] Note: Gas has been suddenly compressed hence its internal energy is used m doing work against external pressure and temperature of gas falls.
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