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A balloon filled with air has a volume of 0.500 L at 25.0 ºC and an internal pressure of 1.20 atm. What will the volume of the balloon be after it is placed in a freezer at –10.0 ºC if the internal pressure does not change?

Respuesta :

Answer:

0.441 L

Explanation:

At constant pressure, Using Charle's law  :-

[tex]\frac {V_1}{T_1}=\frac {V_2}{T_2}[/tex]

Given ,  

V₁ = 0.500 L

V₂ = ?

T₁ = 25.0 °C

T₂ = -10.0 °C  

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

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

T₂ = (-10 + 273.15) K = 263.15 K  

Using above equation as:

[tex]\frac{0.500}{298.15}=\frac{V_2}{263.15}[/tex]

[tex]V_2=\frac{0.500\cdot \:263.15}{298.15}[/tex]

New volume = 0.441 L

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