Question

A spherical balloon with a diameter of 9 m is filled with helium at 20°C and 200 kPa. Determine the mole number and the mass of the helium in the balloon.

Answer 1

Answer:

number of mole is 31342.36 moles

mass is 125.369 kg

Explanation:

Diameter of the spherical balloon d = 9 m

radius r = d/2 = 9/2 = 4.5 m

The volume pf the sphere balloon ca be calculated from

V = $\frac{4}{3} \pi r^3$

V = $\frac{4}{3}* 3.142* 4.5^3$ = 381.75 m^3

Temperature of the gas T = 20 °C = 20 + 273 = 293 K

Pressure of the helium gas = 200 kPa = 200 x 10^3 Pa

number of moles n = ?

Using

PV = nRT

where

P is the pressure of the gas

V is the volume of the gas

n is the mole number of the gas

R is the gas constant = 8.314 m^3⋅Pa⋅K^−1⋅mol^−1

T is the temperature of the gas (must be converted to kelvin K)

substituting values, we have

200 x 10^3 x 381.75 = n x 8.314 x 293

number of moles n  = 76350000/2436 = 31342.36 moles

We recall that n = m/MM

or m = n x MM

where

n is the number of moles

m is the mass of the gas

MM is the molar mass of the gas

For helium, the molar mass = 4 g/mol

substituting values, we have

m = 31342.36 x 4

m = 125369.44 g

m = 125.369 kg

Answer 2

The mole number and the mass of the helium in the balloon are 250.801 kilomoles and 1003.706 kilograms, respectively.

Let suppose that helium contained in the spherical balloon behaves ideally, the mole number ($n$), in kilomoles, is determined by the following expression:

$n = \frac{P\cdot V}{R_{u}\cdot T}$ (1)

Where:

• $P$ - Pressure, in kilopascals.
• $V$ - Volume, in cubic meters.
• $R_{u}$ - Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin.
• $T$ - Temperature, in Kelvin.

The volume and mass of helium ($m$), in kilograms, is described by these two formulas:

$V = \frac{4\pi}{3}\cdot R^{3}$ (2)

$m = n\cdot M$ (3)

Where:

• $R$ - Radius of the sphere, in meters.
• $M$ - Molar mass, in kilograms per kilomole.

If we know that $P = 200\,kPa$, $R = 9\,m$, $R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}$, $T = 293.15\,K$ and $M = 4.002\,\frac{kg}{kmol}$, then the mole number and the mass of the helium in the balloon are:

$V = \frac{4\pi}{3}\cdot (9\,m)^{3}$

$V \approx 3053.628\,m^{3}$

$n = \frac{(200\,kPa)\cdot (3053.628\,m^{3})}{\left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot(293.15\,K)}$

$n = 250.580\,kmol$

$m = (250.801\,kmol)\cdot \left(4.002\,\frac{kg}{kmol} \right)$

$m = 1003.706\,kg$

The mole number and the mass of the helium in the balloon are 250.801 kilomoles and 1003.706 kilograms, respectively.

We kindly invite to check this question on ideal gases: https://brainly.com/question/16211117