A) \[\frac{4}{9}%\]
B) \[\frac{2}{3}%\]
C) 4 %
D) \[\frac{9}{4}%\]
Correct Answer: A
Solution :
[a] \[P{{V}^{\gamma }}=constant\text{ }\left( k \right)\] \[\ell nP+\frac{3}{2}\ell nV=\ell nk\] \[\therefore \frac{\Delta P}{P}+\frac{3}{2}\frac{\Delta V}{V}=0\] \[\therefore \frac{\Delta V}{V}=-\frac{2}{3}\frac{\Delta P}{P}\] \[\therefore \frac{\Delta V}{V}\times 100=-\frac{2}{3}\left( \frac{\Delta P}{P}\times 100 \right)=-\frac{4}{9}\] \[\therefore \] Volume decreases by about \[\frac{4}{9}%\]You need to login to perform this action.
You will be redirected in
3 sec