Dear Avanish
No. of balloons that can be filled = V of H2available/V of one balloon
Calculation of total volume of hydrogen in the cylinder at N.T.P
P1V1/T1 = PV2/T2
P1 = 1 atm P2 = 20 atm
V1 = ? V2 = 2.82 l
T1 = 23 K T2 = 273 + 27 = 300 K
∴ V1 = 20 * 2.82 * 273/300 * 1 = 51.324 l = 51324 ml
Actual volume to be transferred into balloons
= 51324 – 2820 ml = 48504 ml
[∵ 2820 ml of H2 will remain in cylinder]
No. of balloons that can be filled up -= 48504/4851 = 9.999 = 10
Volume of one balloon = 4/3 πr3 = 4/3 * 22/7 * (21/2)3
[∵ r = diameter/2]
= 4851 ml = 4.851L
ALTERNATESOLUTION
Volume of balloon = 4.851 L (as calculate above)
Let no. of balloon to be filled n
∴ Total volume occupied by n balloons = 4.851 * n
Volume of H2 present in cylinder = 2.82 L (given)
∴ Total volume of H2 at NTP = (4.851n + .82)L
P1 = 1 atm P2 = 20 atm
V1 = 4.85 * n + 2.82 L V2 = 2.82 L
T1 = 273 K
P1V1/T1 = P2V2/T2
or 1 * (4.85 * n + 2.82)/273 = 20 * 2.82/300
∴n = 48.504/4.851 = 10
Regards
Arun, (askIITians Forum Expert)
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