Question

A spherical balloon of $21cm$ diameter is to be filled up with ${H_2}$ at NTP from a cylinder containing the gas at $20atm$ and $27^\circ C$. The cylinder can hold $2.82litres$ of water at NTP.
Calculate the number of balloons that can be filled up.
A. $14$
B. $40$
C. $17$
D. $10$

Answer 1

NTP is a concept that is given by IUPAC and it stands for Normal Temperature and Pressure. It is a standard that is used by NIST at a temperature of $20^\circ C$ and has an absolute pressure of $1atm$. The value of density at NTP is $1.204kg/{m^3}$.

Complete step by step answer:
One mole of gas occupies $22.4L$ at NTP.
According to the question; given: the diameter of balloon = $21cm$
So, the radius of balloon = $21/2 = 10.5cm$
Volume of one balloon = $\dfrac{4}{3}\pi {r^3}$ [by using the formula of volume of sphere]
= $\dfrac{4}{3} \times \dfrac{{22}}{7} \times {\left( {\dfrac{{21}}{2}} \right)^3}$= $4851mL$= $4.851L$
Let the balloons be filled = n
Volume of hydrogen covered by balloons = $4.851 \times n$.
It is given that the cylinder has some volume = $2.82litres$
Therefore, total volume held by hydrogen at NTP = $4.851 \times n + 2.82litre$.
At STP (standard temperature and pressure);
${P_1}$= $20atm$
${T_1}$= $293K$
${T_2}$= $300K$
${P_2}$= ?
${V_1}$= $4.851 \times n + 2.82litre$
${V_2}$= $2.82litres$
By using the formula, $\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}$
$1 \times \left( {\dfrac{{4.851n + 2.82}}{{293}}} \right)$= $20 \times \dfrac{{2.82}}{{300}}$
By solving this equation, we get the value of $n$
$\therefore$ the value of $n$ is $10$
Hence, the number of balloons is $10$ that can be filled up.

So, the correct answer is Option D.

Note:
STP [Standard Temperature and Pressure ] and NTP [ Normal Temperature and Pressure] are not same because STP is set by IUPAC at $0^\circ C$and $1bar$ or $100kPa$ whereas NTP is set at $101.325kPa$with the temperature of $20^\circ C$.