A balloon that contains 0.50 L of air at 25 °C is cooled to –196 °C. What volume does the balloon now occupy?

Question

A balloon that contains 0.50 L of air at 25 °C is cooled to –196 °C. What volume does the balloon now occupy?
Analysis
Since this question deals with volume and temperature, Charles’s law is used to determine a final volume because three quantities are known—the initial volume and temperature ([latex]V_{1}[/latex] and [latex]T_{1}[/latex]), and the final temperature ( [latex]T_{2}[/latex]).

Accepted Answer

[1] Identify the known quantities and the desired quantity.

[latex]V_{1}[/latex] = 0.50 L
[latex]T_{1}[/latex]= 25 °C [latex]T_{2}[/latex]= –196 °C [latex]V_{2}[/latex]= ?
known quantities desired quantity

• Both temperatures must be converted to Kelvin temperatures using the equation
K = °C + 273.
• [latex]T_{1}[/latex]= 25 °C + 273 = 298 K
• [latex]T_{2}[/latex]= –196 °C + 273 = 77 K

[2] Write the equation and rearrange it to isolate the desired quantity, [latex]V_{2}[/latex], on one side.
• Use Charles’s law.

[3] Solve the problem.
• Substitute the three known quantities into the equation and solve for [latex]V_{2}[/latex].

• Since the temperature has decreased, the volume of gas must decrease as well.

Question 6.sp.3: A balloon that contains 0.50 L of air at 25 °C is cooled to ...

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