A spherical balloon of $$21\,cm$$  diameter is to be filled with $${H_2}$$  at $$NTP$$  from a cylinder containing the gas at $$20\,atm$$  at $${27^ \circ }C.$$  The cylinder can hold $$2.82\,L$$  of water at $$NTP.$$  The number of balloons that can be filled up is

Solution :
$$\eqalign{ & {\text{The volume of the balloon}} = \frac{4}{3}\pi {r^3} \cr & = \frac{4}{3} \times \frac{{22}}{7} \times {\left( {\frac{{21}}{2}} \right)^3} \cr & = 4851\,mL \cr & {\text{Volume of the cylinder}} = 2820\,mL \cr & {\text{Volume of}}\,{H_2}\,{\text{at }}NTP \cr & = \frac{{20 \times 2820 \times 273}}{{300 \times 1}}mL \cr & = 51324\,mL \cr} $$
After filling the cylinder will have $${H_2}$$  equal to its volume $$ = 2820\,mL$$
$$\eqalign{ & \therefore {\text{Volume of}}\,{H_2}\,{\text{for filling balloons}} \cr & = 51324 - 2820 = 48504\,mL \cr & {\text{Hence no}}{\text{. of balloon to be filled}} \cr & = \frac{{48504}}{{4851}} \approx 10 \cr} $$