Abubble of helium gas has a volume of 0.650 cm3 near the bottom of a large aquarium where the pressure is 1.54 atm and the temperature is 12°c. determine the bubble’s volume upon rising near the top where the pressure is 1.01 atm and 16°c. assume that the number of moles of helium remains constant and that the helium is an ideal gas. (13 pts)

important equations and constants 1 atm = 760 torr = 760 mmhg = 101,325 pa

1ml = 1cm3

pv = nrt

p1v1 = p2v2

v1/t1 = v2/t2

v1/n1 = v2/n2

ptotal = p1 + p2 + p3 + ….

(p1v1)/(n1t1) = (p2v2)/(n2t2)

r = 0.08206 l atm/mol k

1.005 cm³

Explanation:

Using Ideal gas equation for same mole of gas as

Given ,  

V₁ = 0.650 cm³

V₂ = ?

P₁ = 1.54 atm

P₂ = 1.01 atm

T₁ = 12°C

T₂ = 16 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (12 + 273.15) K = 285.15 K  

T₂ = (16 + 273.15) K = 289.15 K  

Using above equation as:

Solving for V₂ , we get:

V₂ = 1.005 cm³

it is important to conserve water because there is very limited amount of freshwater here on earth and if we are not smart about how we use it we will only be runnning out of our supply faster than neccessarry.

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