Asked by Jalea
A perfectly spherical balloon has a radius of 10.0cm inside a cold room where the temperature hovers around 17.0°C, When the balloon is taken outside, its volume increases to 4.50 L. Assuming there is no change in pressure, what would the outside temperature in °C be?
Answers
Answered by
DrBob222
initial volume = (4/3)*pi*r^3 = (4/3)*3.14*(10.0 cm )^3 = ? cc.= V1 in cc but convert to L.
initial T is 17.0 C and that's not "around 17.0 C). Kelvin = 16 + 273 = T1
final T is T2.
Final volume is 4.50 L = V2
Then use P1V1/T1 = P2V2/T2. Substitute from above and solve for T2. That will be in kelvin. Convert to C if that's what you want.
initial T is 17.0 C and that's not "around 17.0 C). Kelvin = 16 + 273 = T1
final T is T2.
Final volume is 4.50 L = V2
Then use P1V1/T1 = P2V2/T2. Substitute from above and solve for T2. That will be in kelvin. Convert to C if that's what you want.
Answered by
DrBob222
Note: Since P1 = P2 you may assume any number and use that for both P1 and P2 OR (and the simplest) is to ignore P1 and P2 in the equation and use V1/T1 = V2/T2
Post your work if you get stuck.
Post your work if you get stuck.
Answered by
Jalea
Super thank youuuuuuuuuuu!!!
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