A weather balloon carries instruments that measure temperature, pressure, and humidity as it rises through the atmosphere. Suppose such a balloon has a volume of 1.2 m^3 at sea level where the pressure is 1 atm and the temperature is 20°C. When the balloon is at an altitude of 11 km (36,000 ft) the pressure is down to 0.5 atm and the temperature is about – 55°C. What is the volume of the balloon then?
18m^3
Well, it seems like our weather balloon is in for some "deflation" as it rises through the atmosphere! Let's calculate the new volume using the ideal gas law. The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the temperatures from Celsius to Kelvin. Adding 273.15 to 20°C gives us a temperature of approximately 293.15 K. And subtracting 273.15 from -55°C gives us a chilly temperature of approximately 218.15 K.
Now, let's set up the equation with the initial conditions:
(1 atm)(1.2 m^3) = n(0.0821 L*atm/(mol*K))(293.15 K)
Solving for n, we find that the number of moles is approximately 0.052 mol.
Now, let's calculate the new volume at the higher altitude using the final conditions:
(0.5 atm)(V-final) = (0.052 mol)(0.0821 L*atm/(mol*K))(218.15 K)
Simplifying the equation, we find that the volume at the higher altitude is approximately 0.286 m^3.
So, it looks like our weather balloon "deflated" a bit, and it now has a volume of approximately 0.286 m^3 at an altitude of 11 km. Safe travels, little balloon!
To solve this problem, we can use the ideal gas law, which states:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin
First, let's convert the given temperatures to Kelvin:
Temperature at sea level: 20°C = 20 + 273.15 = 293.15 K
Temperature at 11 km altitude: -55°C = -55 + 273.15 = 218.15 K
The initial conditions at sea level are:
P1 = 1 atm
V1 = 1.2 m^3
T1 = 293.15 K
The conditions at 11 km altitude are:
P2 = 0.5 atm
V2 = ?
T2 = 218.15 K
Now, let's rearrange the ideal gas law equation to solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
Substituting the given values, we can calculate V2:
V2 = (1 atm * 1.2 m^3 * 218.15 K) / (0.5 atm * 293.15 K)
Simplifying the equation gives us:
V2 = (1.2 * 218.15) / (0.5 * 293.15) m^3
V2 ≈ 0.957 m^3
Therefore, at an altitude of 11 km, the volume of the balloon is approximately 0.957 m^3.
To solve this problem, we can use the ideal gas law, which states:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in m^3)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
We know the initial values at sea level and the final values at an altitude of 11 km. Let's assign the appropriate variables:
Initial conditions:
P1 = 1 atm
V1 = 1.2 m^3
T1 = 20°C = 20 + 273.15 = 293.15 K
Final conditions:
P2 = 0.5 atm
T2 = -55°C = -55 + 273.15 = 218.15 K
We need to find V2, the volume at the given altitude.
First, we need to find n, the number of moles of gas. We can use the ideal gas law to solve this:
n1 = (P1 * V1) / (R * T1)
Substituting the given values:
n1 = (1 atm * 1.2 m^3) / (0.0821 L.atm/mol.K * 293.15 K)
Next, we can find V2 using the same formula:
V2 = (n1 * R * T2) / P2
Substituting the given values and the calculated value of n1:
V2 = ((1 atm * 1.2 m^3) / (0.0821 L.atm/mol.K * 293.15 K)) * (0.0821 L.atm/mol.K * 218.15 K) / (0.5 atm)
Now we can calculate V2.
V2 = (1.2 * 0.0821 * 218.15) / (0.0821 * 293.15 * 0.5) m^3
Simplifying:
V2 ≈ 0.595 m^3
Therefore, the volume of the balloon at an altitude of 11 km is approximately 0.595 m^3.
P*V/T is constant. That is one form of the ideal gas law when the number of moles of gas stays the same.
T must be measured in absolute degrees (K) when solving.
T1 = 293 K. T2 = 218 K
P1*V1/T1 = P2*V2/T2
Solve for V2
V2 = V1*(P1/P2)*(T2/T1)
= V1*(0.5)*(218/293)
You know V1 = 1.2 m^3, so do the numbers.