Question

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?

Answer 1

0.441 L

Step-by-step explanation:

At constant pressure, Using Charle's law :-

`\frac {V_1}{T_1}=\frac {V_2}{T_2}`

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:

`(0.500)/(298.15)=(V_2)/(263.15)`

`V_2=(0.500\cdot \:263.15)/(298.15)`

New volume = 0.441 L