A spherical balloon of 21 cm diameter is to be filled up with hydrogen

1.	A spherical balloon of 21 cm diameter is to be filled up with hydrogen at N.T.P. from acylinder containing the gas at 20 atm. at 27Â°C. If the cylinder can hold 2.82 litre of water, thenumber of balloon that can be filled up are(A) 5(B) 10(C) 20(D) 25
Answer» No. of balloons that can be filled =Vof H2available/Vof one balloon Calculation of total volume of hydrogen in the cylinder at N.T.P P1V1/T1=PV2/T2 P1= 1 atmP2= 20 atm V1= ?V2= 2.82 l T1= 23 KT2= 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 H2will 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 ALTERNATE SOLUTION: Volume of balloon = 4.851 L (as calculate above) Let no. of balloon to be filledn ∴ Total volume occupied bynballoons = 4.851 n Volume of H2present in cylinder = 2.82 L (given) ∴ Total volume of H2at NTP = (4.851n + .82)L P1= 1 atmP2= 20 atm V1= 4.85 n + 2.82 LV2= 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

A spherical balloon of 21 cm diameter is to be filled up with hydrogen at N.T.P. from acylinder containing the gas at 20 atm. at 27Â°C. If the cylinder can hold 2.82 litre of water, thenumber of balloon that can be filled up are(A) 5(B) 10(C) 20(D) 25

Answer»

No. of balloons that can be filled =Vof H2available/Vof one balloon

Calculation of total volume of hydrogen in the cylinder at N.T.P

P1V1/T1=PV2/T2

P1= 1 atmP2= 20 atm

V1= ?V2= 2.82 l

T1= 23 KT2= 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 H2will 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

ALTERNATE SOLUTION:

Volume of balloon = 4.851 L (as calculate above)

Let no. of balloon to be filledn

∴ Total volume occupied bynballoons = 4.851 *n

Volume of H2present in cylinder = 2.82 L (given)

∴ Total volume of H2at NTP = (4.851n + .82)L

P1= 1 atmP2= 20 atm

V1= 4.85 * n + 2.82 LV2= 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