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A spherical balloon of 21 CM diameter to be filled with hydrogen gas at NTP from a cylinder containing a gas containing a gas 20 atm at 300K. The cylinder can hold 2.82 litre of water at ntp .Calculate the number of balloons that can be filled up.

A spherical balloon of 21 CM diameter to be filled with hydrogen gas at NTP from a cylinder containing a gas containing a gas 20 atm at 300K. The cylinder can hold 2.82 litre of water at ntp .Calculate the number of balloons that can be filled up.

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Grade:12

1 Answers

Arun
25750 Points
6 years ago
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)
 
If you find the solution correct then please approve it & in case of any confusion further please do feel free to write me here. I would be more than happy to help further

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