A) 28 atm
B) 68.7 atm
C) 256 atm
D) 8 atm
Correct Answer: C
Solution :
[c] \[{{T}_{1}}=273+27=300K\] \[{{T}_{2}}=273+927=1200K\] For adiabatic process, \[{{P}^{1-\gamma }}{{T}^{\gamma }}=\text{constant}\Rightarrow \text{P}_{1}^{1-\gamma }T_{1}^{\gamma }=\text{P}_{2}^{1-\gamma }T_{2}^{\gamma }\] \[\Rightarrow {{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{1-\gamma }}={{\left( \frac{{{T}_{1}}}{{{T}_{2}}} \right)}^{\gamma }}\Rightarrow {{\left( \frac{{{P}_{1}}}{{{T}_{2}}} \right)}^{1-\gamma }}={{\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)}^{\gamma }}\] \[{{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{1-1.4}}={{\left( \frac{1200}{300} \right)}^{1.4}}\Rightarrow {{\left( \frac{{{P}_{1}}}{{{T}_{2}}} \right)}^{-0.4}}={{\left( 4 \right)}^{1.4}}\] \[{{\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)}^{0.4}}={{4}^{1.4}}\text{or, }{{P}_{2}}={{P}_{1}}{{4}^{\left( \frac{1.4}{0.4} \right)}}={{P}_{1}}{{4}^{\left( \frac{7}{2} \right)}}\] \[={{P}_{1}}\left( {{2}^{7}} \right)=2\times 128=256atm\]
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