# Question de7e1

Nov 9, 2016

270

#### Explanation:

${V}_{1} \to \text{volume of hydrogen in cylinder} = 2.82 L$

${T}_{1} \to \text{Temperature of hydrogen in cylinder}$
$= {27}^{\circ} C = 27 + 273 = 300 K$

${P}_{1} \to \text{Pressure of hydrogen in cylinder} = 20 a t m$

At NTP

${P}_{2} \to \text{Pressure} = 1 a t m$

${T}_{2} \to \text{Temperarture} = 273 K$

V_2->"Volume"=?#

By gas law

$\frac{{P}_{2} {V}_{2}}{T} _ 2 = \frac{{P}_{1} {V}_{1}}{T} _ 1$

$\implies {V}_{2} = \frac{{P}_{1} {V}_{1} {T}_{2}}{{P}_{2} {T}_{1}}$

$= \frac{20 \times 2.82 \times 273}{1 \times 300} = 51.324 L$

Excluding the minlmum remaining 2.82L gas in the cylinder at NTP maximum $\left(51.324 - 2.82\right) L = 48.504 L$ gas can be used for filling baloons.

Volume of each spherical baloon of diameter $7 c m$ is

$= \frac{4}{3} \pi {\left(\frac{7}{2}\right)}^{3} c {m}^{3} = \frac{4}{3} \cdot \frac{22}{7} \cdot {7}^{3} / 8 c {m}^{3}$

$= 0.1796 L$

Hence the maximum number of baloons that can be filled with hydrogen from the cylinder is

$= \text{Total volume of gas"/"volume of a baloon}$

$= \frac{48.504}{0.1796} \approx 270 \text{ }$(rounding to nearest whole number)