Answer:
The minimum volume that the tank must have to inflate this balloon to its 1 m diameter is 174.53 L.
Explanation:
Temperature of the gas = T = 25°C
Pressure of the helium gas in the tank connected to balloon =[tex]P_1=1.5 MPa=1.5\times 10^3 kPa[/tex]
Let the minimum volume of the tank to inflate this balloon to its 1 m diameter be [tex]V_1[/tex]
Temperature of the gas after the valve is opened = T = 25°C
Pressure inside the balloon after the valve is opened =[tex]P_2=500 kPa[/tex]
On opening the valve, helium gas starts moving towards balloon by changing its diameter to 1 m.
Diameter of balloon = d = 1 m
Volume of balloon after the valve is opened = [tex]V_2[/tex]
Radius of the balloon = r = d/2 = (1 m)/2 = 0.5 m
[tex]V_2=\frac{4}{3}\pi r^3=\frac{4}{3}\times 3.14\times (0.5 m)^3[/tex]
[tex]=0.5236 m^3=523 .6 L[/tex]
[tex]1 m^3=1000 L[/tex]
[tex]P_1V_1=P_2V_2[/tex] (Boyle's law)
[tex]V_1=\frac{P_2\times V_2}{P_1}[/tex]
[tex]=\frac{500 kPa\times 523.6 L}{1.5\times 10^3 kPa}=174.53 L[/tex]
The minimum volume that the tank must have to inflate this balloon to its 1 m diameter is 174.53 L.
