Question

A bubble 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

Answer 1

Answer:

1.005 cm³

Explanation:

Using Ideal gas equation for same mole of gas as

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

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:

$\frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}$

$\frac {{1.54}\times {0.650}}{285.15}=\frac {{1.01}\times {V_2}}{289.15}$

Solving for V₂ , we get:

V₂ = 1.005 cm³