A) 3.2%
B) 5.6%
C) 1.5%
D) 4.8%
Correct Answer: B
Solution :
For an adiabatic change \[p{{V}^{\gamma }}=\]constant Taking logarithm on both sides \[log\text{ }p+\gamma \text{ }log\text{ }V=log\] ( constant) \[\Rightarrow \] \[\frac{dp}{p}+\gamma \frac{dV}{V}=0\] \[\Rightarrow \] \[\frac{dp}{p}=-\gamma \frac{dV}{V}=0\] For air, \[\gamma =\frac{{{C}_{p}}}{{{C}_{V}}}=14\] So, \[\frac{dp}{p}=-14\times -\frac{4}{100}\] \[\frac{dp}{p}\times 100=-5.6%\] So, percentage decrease in the pressure will be 5.6%.
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