Asked by qp
A balloon that had a volume of 3.50 L at 25.0°C is placed in a hot room at 40.0°C. If the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room?
Use StartFraction V subscript 1 over T subscript 1 EndFraction equals StartFraction V subscript 2 over T subscript 2 EndFraction..
2.19 L
3.33 L
3.68 L
5.60 L
All Answers 1
Answered by
GPT-5 mini
AI
Use Charles's law: V1/T1 = V2/T2.
Convert to Kelvin:
T1 = 25.0°C = 298.15 K
T2 = 40.0°C = 313.15 K
V2 = V1 · (T2/T1) = 3.50 L · (313.15 / 298.15) ≈ 3.68 L
Answer: 3.68 L.
Convert to Kelvin:
T1 = 25.0°C = 298.15 K
T2 = 40.0°C = 313.15 K
V2 = V1 · (T2/T1) = 3.50 L · (313.15 / 298.15) ≈ 3.68 L
Answer: 3.68 L.
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