a tank contains 26.0 kg of oxygen gas at a gauge pressure of 8.70 atm. if oxygen is replaced by helium, how many kilograms of the latter will be needed to produce a gauge pressure at 7.00 atm?
A gauge pressure of 8.70 atm means the absolute pressure is 9.70 atm. Similarly, a gauge pressur eof 7.00 atm means an absolute pressure of 8.00 atm.
Assuming the temperature remains the same, the gas mass ratio is:
M(He)/M(O2) = (4/32)(8.00/9.70) = 0.1031
So that
M(He) = 26*0.1031 = 2.68 kg
Why did the oxygen and helium go to therapy together? They needed some pressure-release valve-fieing!
To calculate how many kilograms of helium are needed, we can use the ideal gas law:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since the volume and temperature are constant, we can solve for the number of moles using the equation:
P₁V₁ = n₁RT
And then solve for the mass of helium using the equation:
m₂ = n₂M₂
Where:
m₂ = mass of helium
n₂ = number of moles of helium
M₂ = molar mass of helium
Now, let's do the calculations:
Given data for oxygen:
P₁ = 8.70 atm
V₁ = ?
n₁ = ?
R = 0.0821 L·atm/(mol·K)
T = constant (assumed)
Given data for helium:
P₂ = 7.00 atm
V₂ = V₁ (since the volume is constant)
n₂ = ?
M₂ = 4.00 g/mol (molar mass of helium)
First, let's find the number of moles of oxygen (n₁) in the tank:
P₁V₁ = n₁RT
n₁ = (P₁V₁) / (RT)
Now, let's find the number of moles of helium (n₂) needed to reach the desired pressure:
P₂V₂ = n₂RT
n₂ = (P₂V₂) / (RT)
Finally, let's calculate the mass of helium (m₂):
m₂ = n₂M₂
Substituting n₂ with the value we calculated earlier:
m₂ = [(P₂V₂) / (RT)] * M₂
To solve the problem, we can use the ideal gas law equation:
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
Since we are comparing the gauge pressures, the volume (V), ideal gas constant (R), and temperature (T) will remain constant throughout the process. Therefore, we can simplify the equation to:
P1n1 = P2n2
Where:
P1 = initial pressure of the gas (oxygen)
n1 = initial number of moles of gas (oxygen)
P2 = final pressure of the gas (helium)
n2 = final number of moles of gas (helium)
First, let's calculate the initial number of moles of oxygen gas (n1) using the provided information:
Given:
P1 = 8.70 atm
V = unknown
n1 = unknown
R = constant
T = constant
Since we don't know the volume and temperature, we cannot calculate the exact value of n1 using the ideal gas law. However, we can ignore these values since we only need the ratio of n2/n1.
Next, let's calculate the final number of moles of helium gas (n2) using the given information:
Given:
P2 = 7.00 atm
V = unknown
n2 = unknown
R = constant
T = constant
Similarly, we don't know the volume and temperature, but we can still calculate the ratio of n2/n1:
(P1 * n1) / (P2 * n2) = 1
Simplifying, we have:
n2 / n1 = P1 / P2
n2 = (n1 * P2) / P1
Now we can substitute the values and calculate n2:
n2 = (n1 * P2) / P1
= (26.0 kg * 7.00 atm) / 8.70 atm
= 20.93 kg of helium
Therefore, approximately 20.93 kg of helium will be needed to produce a gauge pressure of 7.00 atm.
To solve this problem, we can use Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when temperature remains constant.
According to the problem, the initial pressure of oxygen gas is 8.70 atm, and we want to find out how many kilograms of helium are needed to produce a pressure of 7.00 atm.
First, let's convert the gauge pressures to absolute pressures by adding atmospheric pressure to each. Atmospheric pressure is approximately 1 atm.
Initial oxygen pressure = 8.70 atm + 1 atm = 9.70 atm
Final helium pressure = 7.00 atm + 1 atm = 8.00 atm
Next, we need to determine the initial and final volumes of the gas. Assuming the temperature remains constant, we can write the equation:
(P1 * V1) = (P2 * V2)
Where P1 and P2 are the initial and final pressures, and V1 and V2 are the initial and final volumes.
Since the volume is constant, we can simplify the equation to:
P1 / P2 = V2 / V1
Now, we can plug in the values:
(9.70 atm) / (8.00 atm) = V2 / V1
To find the ratio of the final volume to the initial volume.
Next, we substitute the ratio of the volumes into the density equation to find the mass of helium gas:
ρ = m / V
Where ρ is the density of the gas, m is the mass of the gas, and V is the volume of the gas.
Since density is mass per unit volume, we can rearrange the equation to:
m = ρ * V
To find the mass of helium, we need the density of helium gas. Assuming the temperature remains constant, we can use the ideal gas law:
PV = nRT
Where P is the absolute pressure, V is the volume, n is the number of gas particles (in moles), R is the ideal gas constant, and T is the temperature in Kelvin.
By rearranging the equation, we get:
n = PV / RT
The molar mass of helium is approximately 4 g/mol. So, the density of helium is:
density = (mass of helium) / (volume of helium)
density = (4 g/mol * n) / (V)
Since one mole of gas occupies 22.4 L at standard temperature and pressure (STP), we can rewrite the equation as:
density = (4 g/mol) / (22.4 L/mol) * (n/V)
Finally, multiplying the density and the volume of the gas will give us the mass of helium required:
mass of helium = density * volume
mass of helium = [(4 g/mol) / (22.4 L/mol) * (n/V)] * V
mass of helium = (4 g/mol * n) / 22.4 L
For n, we can use the ideal gas law equation:
n = PV / RT
where P = 8.00 atm (final helium pressure)
V = V2 (final helium volume)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = constant
By substituting the values and solving for n, we can then find the mass of helium:
mass of helium = (4 g/mol * n) / 22.4 L