A partially inflated balloon contains 500cm3 of helium at 27 degree centrigrate and 1atm pressure. What is the volume of helium at altitude of 1800ft where the pressure is 0.5 atm and temperature -30degree centigrate?.
What is the average velocity of the molecules in a sample of oxygen at 100 degree. Hint the mass of an oxygen molecule is 5.3*10^-26kg?
Why did the balloon go to altitude? Because it wanted to show off its gas!
Now, let me calculate the volume of helium at an altitude of 1800ft where the pressure is 0.5 atm and temperature is -30 degrees Celsius.
To solve this, we will use the combined gas equation, which states that P1V1/T1 = P2V2/T2, where P1, V1, and T1 are the initial pressure, volume, and temperature, and P2, V2, and T2 are the final pressure, volume, and temperature.
Given:
P1 = 1 atm
V1 = 500 cm3
T1 = 27 degrees Celsius + 273 = 300 K
P2 = 0.5 atm
T2 = -30 degrees Celsius + 273 = 243 K
Now, we can rearrange the equation to solve for V2:
V2 = (P1 * V1 * T2) / (P2 * T1)
= (1 atm * 500 cm3 * 243 K) / (0.5 atm * 300 K)
= 243000 cm3 / 150 cm3
= 1620 cm3
So, the volume of helium at an altitude of 1800ft will be 1620 cm3. Just remember, it's important to keep the balloon's ego in check!
Now, onto the average velocity of the molecules in a sample of oxygen at 100 degrees.
First, let's convert the temperature to Kelvin:
T = 100 degrees Celsius + 273 = 373 K
The average velocity (v) of the molecules can be calculated using the equation:
v = sqrt((3 * k * T) / m)
Where:
k = Boltzmann's constant (1.38 * 10^-23 J/K)
T = temperature in Kelvin
m = mass of the oxygen molecule (5.3 * 10^-26 kg)
Plugging in the values, we have:
v = sqrt((3 * 1.38 * 10^-23 J/K * 373 K) / 5.3 * 10^-26 kg)
= sqrt(1.6 * 10^-19 J / 5.3 * 10^-26 kg)
= sqrt(3.02 * 10^6 m^2/s^2)
So, the average velocity of the molecules in a sample of oxygen at 100 degrees is approximately 54991 m/s. Just imagine all those molecules zooming around, like tiny race cars on a molecular racetrack!
To solve the first question, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles of gas
R = Ideal gas constant (0.0821 L*atm/(mol*K))
T = Temperature (in Kelvin)
We need to convert the temperatures from Celsius to Kelvin.
To convert Celsius to Kelvin, use the formula:
K = °C + 273.15
To solve the second question, we can use the equation for average velocity:
v = sqrt((3RT) / M)
Where:
v = average velocity
R = Ideal gas constant
T = Temperature (in Kelvin)
M = Mass of the molecule
Let's solve those step-by-step.
Step 1: Convert temperature from Celsius to Kelvin for both questions.
For the first question:
Initial temperature, T1 = 27°C + 273.15 = 300.15 K
Final temperature, T2 = -30°C + 273.15 = 243.15 K
For the second question, temperature = 100°C
T = 100°C + 273.15 = 373.15 K
Step 2: Solve the first question.
Given:
P1 = 1 atm (initial pressure)
V1 = 500 cm³ (initial volume)
P2 = 0.5 atm (final pressure)
Using the ideal gas law equation: PV = nRT
First, solve for n (number of moles):
n = (P1 * V1) / (R * T1)
n = (1 atm * 500 cm³) / (0.0821 L*atm/(mol*K) * 300.15 K)
n ≈ 20.23 moles
Next, solve for the final volume, V2:
V2 = (n * R * T2) / P2
V2 = (20.23 moles * 0.0821 L*atm/(mol*K) * 243.15 K) / 0.5 atm
V2 ≈ 1998.73 cm³
Therefore, the volume of helium at an altitude of 1800 ft, where the pressure is 0.5 atm and temperature is -30°C, is approximately 1998.73 cm³.
Step 3: Solve the second question.
Given:
T = 373.15 K (temperature)
M = 5.3 * 10^-26 kg (mass of oxygen molecule)
Using the average velocity equation: v = sqrt((3RT) / M)
v = sqrt((3 * 0.0821 L*atm/(mol*K) * 373.15 K) / (5.3 * 10^-26 kg))
v ≈ 5.23 * 10^2 m/s
Therefore, the average velocity of oxygen molecules at 100°C is approximately 5.23 * 10^2 m/s.
To find the volume of helium at a different altitude, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional when temperature is held constant.
1. Convert the given temperature and pressures to the appropriate units.
- The initial temperature is 27 degrees Celsius, which is equivalent to 300 Kelvin (K) (Celsius to Kelvin conversion: °C + 273.15).
- The initial pressure is 1 atmosphere (atm).
- The altitude pressure is 0.5 atm.
- The altitude temperature is -30 degrees Celsius, which is equivalent to 243.15 K.
2. Apply Boyle's Law to solve for the new volume.
- Boyle's Law: P1 * V1 = P2 * V2
- P1 = initial pressure (1 atm)
- V1 = initial volume (500 cm³)
- P2 = altitude pressure (0.5 atm)
- V2 = volume at altitude (to be found)
Rearrange the formula to solve for V2:
V2 = (P1 * V1) / P2
Substitute the values into the formula:
V2 = (1 atm * 500 cm³) / 0.5 atm
V2 = 1000 cm³
Therefore, the volume of helium at an altitude of 1800 ft is 1000 cm³.
To find the average velocity of the molecules in a sample of oxygen at 100 degrees Celsius, we can use the root mean square velocity equation from kinetic theory:
v = sqrt((3 * k * T) / m)
where:
- v is the average velocity of the molecules,
- k is the Boltzmann constant (1.38 * 10^-23 J/K),
- T is the temperature in Kelvin (373 K for 100 degrees Celsius),
- m is the mass of an oxygen molecule (5.3 * 10^-26 kg).
1. Convert the given temperature to Kelvin.
- The temperature is 100 degrees Celsius, which is equivalent to 373 Kelvin (Celsius to Kelvin conversion: °C + 273.15).
2. Apply the root mean square velocity equation.
- Substitute the values into the formula:
v = sqrt((3 * (1.38 * 10^-23 J/K) * 373 K) / (5.3 * 10^-26 kg))
v = sqrt(8.067 * 10^(-21) J / 5.3 * 10^-26 kg)
v = sqrt(1.523 * 10^5 m²/s²)
Therefore, the average velocity of the molecules in the sample of oxygen at 100 degrees Celsius is approximately 391 m/s.