We mix 193 grams of oxygen gas with
39 grams of argon gas in a volume of 1140 mL
at 37�C. What will be the final pressure of the gas mixture?
Use PV = nRT and solve for pO2 gas by itself. n = 193/32 = ?
Then PV = nRT and solve for pAr gas by itself. n = 39/atomic mass Ar.
Then ptotal = pO2 + pAr
156.37
To find the final pressure of the gas mixture, we can use the ideal gas law equation which states:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature of the gas (in Kelvin)
First, we need to convert the given temperature from degree Celsius (°C) to Kelvin (K).
T(K) = T(°C) + 273.15
T(K) = 37 + 273.15
T(K) = 310.15 K
Next, we need to calculate the number of moles of each gas using their respective molecular weights (grams/mole).
For oxygen (O2):
Molecular weight of O2 = 32 g/mol
Number of moles of O2 = mass of O2 (g) / molecular weight of O2 (g/mol)
Number of moles of O2 = 193 g / 32 g/mol
Number of moles of O2 = 6.03 mol
For argon (Ar):
Molecular weight of Ar = 39.95 g/mol
Number of moles of Ar = mass of Ar (g) / molecular weight of Ar (g/mol)
Number of moles of Ar = 39 g / 39.95 g/mol
Number of moles of Ar = 0.98 mol
Now, we can calculate the total number of moles of gas in the mixture by summing up the moles of each gas.
Total number of moles = moles of O2 + moles of Ar
Total number of moles = 6.03 mol + 0.98 mol
Total number of moles = 7.01 mol
Next, we need to convert the given volume from milliliters (mL) to liters (L).
V(L) = V(mL) / 1000
V(L) = 1140 mL / 1000
V(L) = 1.14 L
Finally, we can substitute the values into the ideal gas law equation and solve for the final pressure (P).
P * V = n * R * T
P = (n * R * T) / V
P = (7.01 mol * 0.0821 L•atm/mol•K * 310.15 K) / 1.14 L
P = 17.08 atm
Therefore, the final pressure of the gas mixture is approximately 17.08 atm.
To find the final pressure of the gas mixture, we need to apply the ideal gas law equation: PV = nRT.
First, let's find the number of moles for each gas using their given masses and molar masses.
For oxygen gas (O2):
Mass of oxygen gas (m) = 193 grams
Molar mass of oxygen (M) = 32 g/mol (16 g/mol × 2)
Number of moles of oxygen (n) = m / M
For argon gas (Ar):
Mass of argon gas (m) = 39 grams
Molar mass of argon (M) = 40 g/mol
Number of moles of argon (n) = m / M
Next, let's convert the given volume from milliliters (mL) to liters (L) since the ideal gas law requires the volume in liters.
Volume of gas (V) = 1140 mL
Convert V to L: V = 1140 mL ÷ 1000 mL/L = 1.14 L
Now, let's substitute the values into the ideal gas law equation:
For oxygen: (PV = nRT)
PO2 × V = nO2 × RT
PO2 = (nO2 × RT) / V
For argon: (PV = nRT)
Par × V = nAr × RT
Par = (nAr × RT) / V
Since we're looking for the final pressure of the gas mixture, the total pressure (Ptotal) is simply the sum of the partial pressures of each gas:
Ptotal = PO2 + Par
Now, let's calculate the partial pressures:
For oxygen:
nO2 = (mass of oxygen gas) / (molar mass of oxygen)
nO2 = 193 g / 32 g/mol
nO2 ≈ 6.03 mol
For argon:
nAr = (mass of argon gas) / (molar mass of argon)
nAr = 39 g / 40 g/mol
nAr = 0.975 mol
Now, let's substitute the values into the equations:
PO2 = (nO2 × RT) / V
PO2 = (6.03 mol × 0.0821 L·atm/(mol·K) × 310 K) / 1.14 L
PO2 ≈ 16.43 atm
Par = (nAr × RT) / V
Par = (0.975 mol × 0.0821 L·atm/(mol·K) × 310 K) / 1.14 L
Par ≈ 0.211 atm
Finally, let's calculate the total pressure of the gas mixture:
Ptotal = PO2 + Par
Ptotal ≈ 16.43 atm + 0.211 atm
Ptotal ≈ 16.64 atm
Therefore, the final pressure of the gas mixture will be approximately 16.64 atm.