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A balloon with the volume of 24.0 L and at 23 °C and pressure of 1.00 atm is let go. The balloon after 20 minutes reaches a much higher elevation where due to lower pressure of 0.500 atm is now expands to 28.0 liters. What is the temperature at that elevation in Celsius degrees?  [Hint: combined gas law requires temperature input in kelvin and produces results also in kelvin].

Question

A balloon with the volume of 24.0 L and at 23 °C and pressure of 1.00 atm is let go. The balloon after 20 minutes reaches a much higher elevation where due to lower pressure of 0.500 atm is now expands to 28.0 liters. What is the temperature at that elevation in Celsius degrees?  [Hint: combined gas law requires temperature input in kelvin and produces results also in kelvin].

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Solution

To solve this problem, we will use the Combined Gas Law which states that the ratio of the product of pressure and volume to temperature is constant for a given amount of gas. The formula is:

P1V1/T1 = P2V2/T2

Where: P1 = initial pressure = 1.00 atm V1 = initial volume = 24.0 L T1 = initial temperature = 23 °C = 23 + 273.15 = 296.15 K (converted to Kelvin)

P2 = final pressure = 0.500 atm V2 = final volume = 28.0 L T2 = final temperature = ? (this is what we need to find)

Substituting the known values into the formula, we get:

(1.00 atm * 24.0 L) / 296.15 K = (0.500 atm * 28.0 L) / T2

Solving for T2, we get:

T2 = (0.500 atm * 28.0 L) * 296.15 K / (1.00 atm * 24.0 L)

T2 = 370.19 K

However, the question asks for the temperature in Celsius, so we need to convert this back from Kelvin to Celsius:

T2 = 370.19 K - 273.15 = 97.04 °C

So, the temperature at the higher elevation is approximately 97.04 °C.

This problem has been solved

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