If 12.0g of molecular oxygen, O2, is required to inflate a balloon to a certain size at 27 degree celcius, what mass of molecular oxygen required to inflate it to the same size and pressure at 81 degree celcius?
Convert 12.0 g oxygen to moles and use PV=nRT to solve for n.
Then use PV-nRT a second time but substitute the nhew temperature. You can make up values for P and V (I would use 1 and 1) and use the same numbers for both sets of values since the P and V don't change.
To solve this problem, we can use the ideal gas law, which states:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin
First, let's find the initial number of moles of O2 at 27 degrees Celsius.
Given:
Temperature at 27 degrees Celsius = 27 °C
To convert temperature from Celsius to Kelvin, we use the equation:
T(K) = T(°C) + 273.15
T(27 °C) = 27 °C + 273.15 = 300.15 K
Next, we need to convert the mass of the initial O2 sample to moles.
Given:
Mass of O2 = 12.0 g
Molar mass of O2 = 32 g/mol
Number of moles = Mass / Molar mass
Number of moles = 12.0 g / 32 g/mol ≈ 0.375 mol
Now, let's calculate the final mass of O2 required to inflate the balloon to the same size and pressure at 81 degrees Celsius.
Given:
Temperature at 81 degrees Celsius = 81 °C
T(81 °C) = 81 °C + 273.15 = 354.15 K
Now, we know that the pressure, volume, and temperature are constant between the two scenarios (same size and pressure). Therefore, we can set up the following equation:
(P * V) / T = n
Since the pressure, volume, and number of moles are constant, we can write:
P1 * V1 / T1 = P2 * V2 / T2
Plugging in the known values, we have:
P1 * V1 / T1 = P2 * V2 / T2
(T2 / T1) = (P2 * V2) / (P1 * V1)
(T2 / T1) = (n2 * R) / (n1 * R)
(T2 / T1) = n2 / n1
Therefore:
n2 = n1 * (T2 / T1)
n2 = 0.375 mol * (354.15 K / 300.15 K)
n2 ≈ 0.443 mol
Now, we can convert the final number of moles to mass using the molar mass of oxygen.
Mass = Number of moles * Molar mass
Mass = 0.443 mol * 32 g/mol ≈ 14.2 g
Therefore, approximately 14.2 grams of molecular oxygen are required to inflate the balloon to the same size and pressure at 81 degrees Celsius.
To find the mass of molecular oxygen required to inflate the balloon at 81 degrees Celsius, we need to use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
The ideal gas constant (R) is a constant value that is equal to 0.0821 L·atm/mol·K.
Let's assume the pressure and volume are constant for both situations. Therefore, we can write the equation as:
n₁T₁ = n₂T₂
Where:
n₁ = number of moles at 27°C
n₂ = number of moles at 81°C
T₁ = temperature in Kelvin at 27°C (27°C + 273.15 = 300.15 K)
T₂ = temperature in Kelvin at 81°C (81°C + 273.15 = 354.15 K)
Now, let's calculate the number of moles of O2 at 27°C using the given mass.
Given:
Mass of O2 = 12.0 grams
To convert mass to moles, we need to know the molar mass of O2. The molar mass of O2 is approximately 32.0 g/mol since oxygen has an atomic mass of about 16.0 g/mol.
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 12.0 g / 32.0 g/mol
Moles of O2 ≈ 0.375 mol
Now that we know the number of moles at 27°C, let's calculate the number of moles at 81°C using the equation n₁T₁ = n₂T₂.
n₂ = (n₁ * T₁) / T₂
n₂ = (0.375 mol * 300.15 K) / 354.15 K
n₂ ≈ 0.318 mol
To find the mass of O2 at 81°C, multiply the number of moles by the molar mass of O2:
Mass of O2 at 81°C = n₂ * Molar mass of O2
Mass of O2 at 81°C = 0.318 mol * 32.0 g/mol
Mass of O2 at 81°C ≈ 10.2 grams
Therefore, approximately 10.2 grams of molecular oxygen is required to inflate the balloon to the same size and pressure at 81 degrees Celsius.