Asked by Monica
In an air-conditioned room at 19.0C , a spherical balloon had the diameter of 50.0cm . When taken outside on a hot summer day, the balloon expanded to 51.0cm in diameter. What was the temperature outside? Assume that the balloon is a perfect sphere and that the pressure and number of moles of air molecules remains the same.
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Answered by
Writeacher
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Answered by
drwls
Remember and apply the perfect gas law:
P V = n R T
In this case, pressure (P) and the number of moles (n) are constant. The volume (V) is therefore proportional to the absolute temerature, T.
In this case,
V2/V1 = (R2/R1)^3 = 1.0612 = T2/T1
T1 (the initial temperature) = 19.0 C = 292.2 K
Therefore T2 = 310.1 K = 36.9 C
P V = n R T
In this case, pressure (P) and the number of moles (n) are constant. The volume (V) is therefore proportional to the absolute temerature, T.
In this case,
V2/V1 = (R2/R1)^3 = 1.0612 = T2/T1
T1 (the initial temperature) = 19.0 C = 292.2 K
Therefore T2 = 310.1 K = 36.9 C
Answered by
Mr.Hojman
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