A tank contains 30.0 kg of oxygen at a gauge pressure of 8.7atm. If the oxygen is replaced by helium at the same temperature, how many kilograms of helium will be needed to produce a gauge pressure of 7.0atm

To solve the problem of replacing oxygen with helium in a tank while adjusting for different pressures, we can use the ideal gas law and the relationship between pressure, volume, and the number of moles of gas. Let's break this down step by step.

Understanding the Ideal Gas Law

The ideal gas law is expressed as:

PV = nRT

Where:

• P = pressure (in atmospheres)
• V = volume (in liters)
• n = number of moles of gas
• R = ideal gas constant (0.0821 L·atm/(K·mol))
• T = temperature (in Kelvin)

Initial Conditions with Oxygen

We start with 30.0 kg of oxygen. The molar mass of oxygen (O₂) is approximately 32.0 g/mol. To find the number of moles of oxygen, we convert kilograms to grams:

30.0 kg = 30,000 g

Now, we calculate the number of moles:

n_O2 = 30,000 g / 32.0 g/mol = 937.5 moles

Calculating the Volume of the Tank

Using the gauge pressure of oxygen, which is 8.7 atm, we can find the volume of the tank. Since gauge pressure is the pressure above atmospheric pressure, the absolute pressure (P_abs) is:

P_abs = 8.7 atm + 1 atm = 9.7 atm

Now, we can rearrange the ideal gas law to solve for volume (V):

V = nRT / P

Assuming the temperature remains constant, we can use the number of moles of oxygen:

V = (937.5 moles) * (0.0821 L·atm/(K·mol)) * T / 9.7 atm

Since we don't have the temperature, we will keep it as a variable (T) for now.

Replacing Oxygen with Helium

Next, we need to find out how many kilograms of helium will be needed to achieve a gauge pressure of 7.0 atm. The absolute pressure for helium will be:

P_abs = 7.0 atm + 1 atm = 8.0 atm

Finding the Number of Moles of Helium

Using the same volume (V) calculated earlier, we can express the number of moles of helium (n_He):

n_He = PV / RT

Substituting the values:

n_He = (8.0 atm) * V / (0.0821 L·atm/(K·mol) * T)

Setting the Volumes Equal

Since the volume of the tank remains constant, we can set the two equations for volume equal to each other:

(937.5 moles * 0.0821 * T) / 9.7 = (n_He * 0.0821 * T) / 8.0

Notice that T and R cancel out, simplifying our equation:

937.5 / 9.7 = n_He / 8.0

Now, we can solve for n_He:

n_He = (937.5 * 8.0) / 9.7

n_He ≈ 773.0 moles

Calculating the Mass of Helium

The molar mass of helium (He) is about 4.0 g/mol. To find the mass of helium needed:

mass_He = n_He * molar mass_He

mass_He = 773.0 moles * 4.0 g/mol = 3092 g

Converting grams to kilograms:

mass_He = 3092 g / 1000 = 3.092 kg

Final Result

To achieve a gauge pressure of 7.0 atm in the tank, approximately 3.09 kg of helium will be required. This calculation illustrates how gas properties and the ideal gas law can help us understand the relationships between pressure, volume, and mass in different scenarios.