An inflated balloon has a volume of 5.75 L at a temperature of 22°C. At what temperature in degrees Celsius will the volume of the balloon decrease to 5.38 L?

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An inflated balloon has a volume of 5.75 L at a temperature of 22°C. At what temperature in degrees Celsius will the volume of the balloon decrease to 5.38 L?

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Answer 1 · 2016-03-02T01:32:57

The volume will be at $5.38 L$ $"5.38 L"$ when the temperature is at ${2.85}^{\circ} C$ $color(blue)("2.85"^@"C"")$ .

Explanation:

This is an example of Charles' law which states that at constant pressure, the volume is directly proportional to the temperature in Kelvins. The equation to use is $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1=V_2/T_2$ .

Given/Known
$V_{1} = 5.75 L$ $V_1="5.75 L"$
$T_{1} = 22^{\circ} C +273.15=295 K$ $T_1="22"^@"C""+273.15=295 K"$
$V_{2} = 5.38 L$ $V_2="5.38 L"$

Unknown
$T_{2}$ $T_2$

Solution
Rearrange the equation to isolate $T_{2}$ $T_2$ . Substitute the known values into the equation and solve. Convert the temperature from Kelvins to degrees Celsius.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1=V_2/T_2$

$T_{2} = \frac{T_{1} V_{2}}{V_{1}}$ $T_2=(T_1V_2)/V_1$

$T_2=((295"K")*(5.38cancel"L"))/(5.75cancel"L")$

$T_2="276 K"$

Convert Kelvins to $""^@"C""$ .

$""^@"C""="K-273.15$

$""^@"C"=276-273.15="2.85"^@"C"$

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An inflated balloon has a volume of 5.75 L at a temperature of 22°C. At what temperature in degrees Celsius will the volume of the balloon decrease to 5.38 L?

1 Answer

Explanation:

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