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 \[\frac{{{T}^{\gamma }}}{{{P}^{\gamma -1}}}=\text{constant}\] where\[\gamma \] is ratio of specific heats. Given, \[{{T}_{1}}=27{{\,}^{o}}C=27+273=300\,K,{{P}_{1}}=P\] \[{{P}_{2}}=\frac{P}{8},\gamma =\frac{5}{3}\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{P}_{1}}}{{{P}_{2}}} \right)}^{\frac{\gamma -1}{\gamma }}}\] \[\Rightarrow \] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{8}{1} \right)}^{\frac{\frac{5-1}{3}}{5/3}}}\] \[={{(8)}^{0.4}}\] \[=2.297\] \[\Rightarrow \] \[{{T}_{2}}=\frac{{{T}_{1}}}{2.297}=\frac{300}{2.297}\] \[=130.6\,K\simeq 131K\] \[\Rightarrow \] \[{{T}_{2}}=131-273\] \[=-142{{\,}^{o}}C\] Note: Gas has been suddenly compressed hence its internal energy is used in doing work against external pressure and temperature of gas falls.
You need to login to perform this action.
You will be redirected in
3 sec
