Question

Let the initial pressure in bulb ( A=P ) and Final pressure ( =mathrm{P}_{2} ) Final pressure is ( 40 % ) of the initial pressure. So ( mathrm{P}_{2} ) ( =(40 / 100) mathrm{P} )
Volume in ( A=100 mathrm{mL} )
Let initially volume of ( mathrm{B}=mathrm{V} ) After opening the stop cock final volume will be ( (100+mathrm{V}) mathrm{mL} )
According to Boyle's law ( mathrm{P}_{1} mathrm{V}_{1}=mathrm{P}_{2} mathrm{V}_{2} )
( mathrm{P} times 100=(40 / 100) mathrm{P} times(100+mathrm{V}) )
( 100=40+(40 / 100) vee )
( 100-40=(40 / 100) vee )
( (60 times 100) / 40=V )
( V=150 mathrm{mL} )
Thus, the volume of bulb ( B=150 mathrm{mL} )

# 8. Two gas bulbs A and B are connected by a tube having a stopcock. Both A has a volume of 100 mL and contains hydrogen. After opening the gas from A to the evacuated bulb B, the pressure falls down by 40%. The volume (mL) of B must be (a) 75 (b) 150 (c) 125 (d) 250

Solution