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A sealed balloon is filled with 1.30 L of helium at 17°C and 1.03 atm. The balloon rises to a point in the atmosphere where the pressure is 225 torr and the temperature is −31 °C. What is the change in the volume of the balloon as it ascends from 1.03 atm to a pressure of 225 torr?

Respuesta :

Answer:

ΔV=2.481 L

Explanation:

As we know that

[tex]\frac{PV}{T}=\frac{P_{2}V_{2} }{T_{2}}[/tex]

Always use temperature in Kelvin and Pressure in Atm

Given Data

P₁=1.03 atm

P₂=225 torr =(225/760)=0.296 atm

V₁=1.30

T=17⁰C=290.15 K

T₂=-31⁰C=242.15 K

Substitute the given values

So

[tex]\frac{1.03*1.30}{290} =\frac{V_{2}(0.296) }{242.15} \\1.2224*10^{-3}V_{2}=4.6217*10^{-3}\\ V_{2}=3.781[/tex]

You have to find the difference of two volumes:

So

ΔV=3.781 - 1.30

ΔV=2.481 L

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