Question

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

• A. 11
• B. 10
• C. 8
• D. 1

Answer 1

The correct option is B 10

The total volume of hydrogen in the cylinder at N.T.P can be calculated as:

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Here,

$P_1$ = 1 atm $P_2$ = 20 atm

and $V_2$= 2.82 l

$T_1$ = 23 K $T_2$ = 273 + 27 = 300 K

∴ $V_1$ = 20 * 2.82 * 273/300 * 1 = 51.324 l = 51324 ml.

Actual volume to be transferred into balloons

= 51324 – 2820 ml = 48504 ml

Number of balloons that can be filled up,$\frac{48504}{4850}$ = 9.99 =10

The Volume of one balloon can be calculated as

$\frac{4}{3}\pi r^3\frac{4}{3}\times \frac{22}{7}\times {\left(\frac{21}{2}\right)}^3$

= 4851 ml = 4.851L

by using this volume value,

The volume of balloon = 4.851 L

taking a number of balloons to be filled = n

∴ Total volume occupied by n balloons = $4.185\times 5$

Volume of $H_2$ present in cylinder = 2.82 L (given)

∴ Total volume of$H_2$ at NTP = (4.851n + .82)L

$P_2$= 20atm $P_1$= 1atm

V1 = 4.85$\times$ n + 2.82 L$V_2$ = 2.82 L

T1 = 273 K

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

∴ $n$ = 48.504/4.851 =$10$

Hence, 10 is the answer.