Solution:
Solution:
Here, $$ V_1 = 0.1\, m^3, P_1 $$ = 1 atm = 76 cm of Hg
$$V_2 = 0.1 + 0.09 = 0.19\, m^3$$
Assuming temperature to remain constant,
$$P_1V_1= P_2V_2$$
or $$P_2 = \frac{P_1 V_1}{V_2}$$
$$ = \frac{0.1}{0.19 } \times $$ 76 cm of Hg
=40 cm of Hg
$$\therefore$$ Change in pressure= (76 - 40) cm of Hg
= 36 cm or 0.36 m of Hg.
