A) 227.36 K
B) 500.30 K
C) 454.76 K
D) \[-47{}^\circ C\]
Correct Answer: A
Solution :
For an adiabatic process \[{{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{2}^{\gamma -1}\] \[\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)={{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\gamma -1}}\] \[\Rightarrow \] \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)}^{\gamma -1}}\] \[\frac{300}{{{T}_{2}}}={{(2)}^{\gamma -1}}\] \[\frac{300}{{{T}_{2}}}={{(2)}^{1.4-1}}\] \[{{T}_{2}}=227.36K\]You need to login to perform this action.
You will be redirected in
3 sec