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a balloon with an initial volume of 3.4 l and an initial temp of 35 C deflates to 2.8l. what is the new temperature?

User Humilton
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Final answer:

Using Charles's Law, the new temperature of a balloon deflating from 3.4 L at 35°C to 2.8 L is calculated to be approximately -20.21°C.

Step-by-step explanation:

The question pertains to the Charles's Law, which states that for a given amount of gas at constant pressure, the volume of the gas is directly proportional to its temperature (measured in Kelvins). To find the new temperature of a balloon that deflates from an initial volume of 3.4 liters at 35°C (308.15 K) to a new volume of 2.8 liters, we can use the formula:

V1/T1 = V2/T2

Converting the initial temperature to Kelvins:

T1 = 35°C + 273.15 = 308.15 K

Now, rearrange the formula to solve for the new temperature (T2):

T2 = (V2 × T1) / V1 = (2.8 L × 308.15 K) / 3.4 L

Calculate T2:

T2 = (2.8 L × 308.15 K) / 3.4 L \approx 252.94 K

Converting back to Celsius:

T2 ≈ 252.94 K - 273.15 = -20.21°C

So, the new temperature of the balloon is approximately -20.21°C.

User Chamly Idunil
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