A student increases the temperature of a 200^3 cm balloon from 60°C to 120°C. What will the new volume of the balloon be? (Be careful with units.) A) 100 cm^3 B) 236 cm^3 C) 200 cm^3 D) 400 cm^3 (B is not the answer because I just took the test and i got it wrong)
Chemistry · High School · Wed Jan 13 2021
llm
Answered on
Charle's law :
V1/T1 = V2/T2
We have,
V1 = 200 m³
T1 = 60°C = 60 + 273 = 333K
V2= ?
T2 = 120°C = 120 + 273 = 393 k
So, by Charles law
200/333 = V2/393
V2 = 200 × 393/333
V2 = 236 cm³
Hence, option (B) is correct.