A) Lesss than P
B) More than P
C) P
D) Either (a) or (c)
Correct Answer: B
Solution :
\[P{{V}^{\gamma }}=\]constant Þ \[\frac{{{P}_{2}}}{{{P}_{1}}}={{\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)}^{\gamma }}={{\left( \frac{{{V}_{1}}}{{{V}_{1}}/4} \right)}^{\gamma }}={{4}^{\gamma }}\] Þ \[{{P}_{2}}={{4}^{\gamma }}P\] As g is always greater than one so \[{{4}^{\gamma }}>4\] Þ \[{{P}_{2}}>4P\]
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