A weather balloon is filled with helium that occupies a volume of 5.00 x 104L At 0.995 atm and 32.0℃. After it is released, it rises to a location where the pressure is 0.720 atm and the temperature is -12.0℃. What is the volume of the balloon at the new location?
since PV/T is constant, you want V such that
0.720V/(-12.0+273.15) = 0.995*5.00*10^4/(32.0+273.15)
V = 59134 L
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, we need to convert the given temperatures from Celsius to Kelvin:
32.0℃ + 273.15 = 305.15 K (initial temperature)
-12.0℃ + 273.15 = 261.15 K (new temperature)
Next, let's calculate the number of moles at both locations using the ideal gas law equation.
For the initial location:
P1 = 0.995 atm
V1 = 5.00 x 10^4 L
T1 = 305.15 K
For the new location:
P2 = 0.720 atm
V2 = ?
T2 = 261.15 K
Now we can set up two equations using the ideal gas law at the initial and new locations:
P1V1 = nRT1
P2V2 = nRT2
Since we have the same number of moles (n) at both locations, we can set both equations equal to each other:
P1V1/T1 = P2V2/T2
Rearranging the equation to solve for V2, we get:
V2 = (P1V1 * T2) / (P2 * T1)
Now we can substitute the given values into the equation and solve:
V2 = (0.995 atm * 5.00 x 10^4 L * 261.15 K) / (0.720 atm * 305.15 K)
Calculating this expression will give us the volume of the balloon at the new location.
To solve this problem, we can use the combined gas law equation, which relates the pressure, volume, and temperature of a gas:
(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.
We are given the following values:
P₁ = 0.995 atm
V₁ = 5.00 x 10⁴ L
T₁ = 32.0 ℃ or 305.15 K
P₂ = 0.720 atm
T₂ = -12.0 ℃ or 261.15 K
Now, let's substitute these values into the equation and solve for V₂:
(0.995 atm × 5.00 x 10⁴ L) / (305.15 K) = (0.720 atm × V₂) / (261.15 K)
Now we rearrange the equation to solve for V₂:
V₂ = (0.995 atm × 5.00 x 10⁴ L × 261.15 K) / (0.720 atm × 305.15 K)
Simplifying the equation:
V₂ = (0.995 atm × 261.15 K × 5.00 x 10⁴ L) / (0.720 atm × 305.15 K)
V₂ ≈ 1.228 x 10⁴ L
Therefore, the volume of the balloon at the new location is approximately 1.228 x 10⁴ L.