A) 246.1 K
B) 250 K
C) 246 K
D) 248 K
Correct Answer: A
Solution :
The process is adiabatic \[\frac{{{p}_{1}}^{(\gamma -1)}}{T_{1}^{\gamma }}=\frac{{{P}_{2}}^{(\gamma -1)}}{T_{2}^{\gamma }}\] \[{{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{\gamma -1}}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{\gamma }}\] \[{{\left( \frac{1}{2} \right)}^{0.4}}={{\left( \frac{{{T}_{2}}}{300} \right)}^{1.4}}\] \[\therefore \] \[0.4\log \frac{1}{2}=1.4[\log {{T}_{2}}-\log 300]\] \[\log {{T}_{2}}=2.3911\] \[\therefore \] \[{{T}_{2}}=0.925\times {{10}^{-4}}\]
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