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?
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.
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.
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Well, I must say, that balloon really knows how to chill! Now let's see if we can figure out the temperature outside. First, we need to convert the volume from liters to cubic centimeters. Since 1 liter is equal to 1000 cubic centimeters, 4.50 liters would be 4500 cubic centimeters.
Now, we have the initial volume of the balloon, which is given by the formula V = (4/3)πr^3. Plugging in the radius of 10.0 cm, we can solve for the initial volume. However, we need to be careful here because the volume will increase when the balloon gets outside!
So, we have the initial volume as V1 = (4/3)π(10.0 cm)^3 and the final volume as V2 = 4500 cm^3. Since the pressure remains constant, we can set up an equation: V1 / V2 = T1 / T2, where T1 represents the initial temperature in Celsius and T2 is the unknown outside temperature.
Now, we can solve for T2 using some algebra: T2 = (V2 / V1) * T1. Plugging in the given values, we get T2 = (4500 cm^3) / [(4/3)π(10.0 cm)^3] * 17.0°C.
Calculating this out, we get T2 ≈ -189.1°C. Whoa, that's ice-cold! So, it seems like the outside temperature would be around -189.1°C. Time to bring a warm coat and gloves if you're heading out there! Stay warm!
To solve this problem, we can use the combined gas law, which is derived from the ideal gas law and is used to relate the changes in pressure, volume, and temperature of a gas.
The combined gas law is expressed as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where P1 and P2 are the initial and final pressures, V1 and V2 are the initial and final volumes, T1 and T2 are the initial and final temperatures.
In this case, we are given the initial and final volumes (V1 = volume inside the cold room, and V2 = volume outside), the initial temperature (T1 = temperature inside the cold room), and we are told that there is no change in pressure.
Since there is no change in pressure, we can cancel it out from the equation, simplifying it to:
(V1) / (T1) = (V2) / (T2)
Plugging in the known values, we have:
(4.50 L) / (T1) = (10.0 cm)^3 / (T2)
To solve for T2, we need to convert the radius of the balloon from cm to L by using the volume formula for a sphere (V = (4/3) * π * r^3):
V1 = (4/3) * π * (10.0 cm)^3
V1 = (4/3) * π * (10.0 cm)^3 * (1 L / 1000 cm^3)
V1 = (4/3) * π * (10.0 cm)^3 * (1 L / 1000 cm^3)
V1 ≈ 4.189 L
Now we can rewrite the equation as:
4.189 L / (T1) = 4.50 L / (T2)
To isolate T2, we can cross-multiply:
4.189 L * (T2) = 4.50 L * (T1)
Now we can solve for T2:
T2 = (4.50 L * T1) / 4.189 L
Substituting T1 = 17.0°C:
T2 = (4.50 L * 17.0°C) / 4.189 L
T2 ≈ 18.16°C
Therefore, the outside temperature is approximately 18.16°C.