Aparticular balloon can be stretched to a maximum surface area of 1257 cm2. the balloon is filled with 3.1 l of helium gas at a pressure of 754 torr and a temperature of 294 k . the balloon is then allowed to rise in the atmosphere. assume an atmospheric temperature of 273 k and determine at what pressure the balloon will burst. (assume the balloon to be in the shape of a sphere.)

P = 0.6815 atm

Explanation:

Pressure = 754 torr

The conversion of P(torr) to P(atm) is shown below:

So,  

Pressure = 754 / 760 atm = 0.9921 atm

Temperature = 294 K

Volume = 3.1 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9921 atm × 3.1 L = n × 0.0821 L.atm/K.mol × 294 K  

⇒n of helium gas= 0.1274 moles

Surface are = 1257 cm²

For a sphere, Surface area = 4 × π × r² = 1257 cm²

r² = 1257 / 4 × π ≅ 100 cm²

r = 10 cm

The volume of the sphere is :

Where, V is the volume

r is the radius

V = 4190.4762 cm³

1 cm³ = 0.001 L

So, V (max) = 4.19 L

T = 273 K

n = 0.1274 moles

Using ideal gas equation as:

PV=nRT

Applying the equation as:

P × 4.19 L = 0.1274 × 0.0821 L.atm/K.mol × 273 K  

P = 0.6815 atm

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

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