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A binary mixture of oxygen and nitrogen with partial pressures in the ratio 0.21 and 0.79 is contained in a vessel at 300 K. If the total pressure of the mixture is 1 * 10 5 N/m2, find molar concentration of oxygen
- a)8.42 * 10 -1 kg mol/m3
- b)8.42 * 10 -2 kg mol/m3
- c)8.42 * 10 -3 kg mol/m3
- d)8.42 * 10 -4 kg mol/m3
Correct answer is option 'C'. Can you explain this answer?
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A binary mixture of oxygen and nitrogen with partial pressures in the ...
Explanation: 0.21 * 10 5/8341 * 300 = 8.42 * 10 -3 kg mol/m3.
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A binary mixture of oxygen and nitrogen with partial pressures in the ...
To find the molar concentration of oxygen, we need to use the partial pressure of oxygen, the total pressure of the mixture, and the ideal gas law equation.
Given:
Partial pressure of oxygen (P_oxygen) = 0.21 * (1 * 10^5 N/m^2) = 2.1 * 10^4 N/m^2
Partial pressure of nitrogen (P_nitrogen) = 0.79 * (1 * 10^5 N/m^2) = 7.9 * 10^4 N/m^2
Total pressure of the mixture (P_total) = 1 * 10^5 N/m^2
Temperature (T) = 300 K
The ideal gas law equation is given as:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant
T = temperature
We can rearrange the equation to solve for the moles of gas:
n = PV / RT
Now, let's calculate the moles of oxygen:
n_oxygen = P_oxygen * V / RT
Since we are looking for the molar concentration, we need to divide the moles of oxygen by the volume. The molar concentration is given in units of kg mol/m^3. To convert the molar concentration from mol/m^3 to kg mol/m^3, we need to divide by the molar mass of oxygen.
The molar mass of oxygen (M_oxygen) is approximately 32 g/mol.
To convert the molar concentration to kg mol/m^3, we use the following conversion factor:
1 mol/m^3 = (1/1000) kg mol/m^3
Therefore, the molar concentration of oxygen (C_oxygen) is given by:
C_oxygen = (n_oxygen / V) * (1/M_oxygen) * (1/1000)
Now, let's substitute the values into the equation:
C_oxygen = (P_oxygen * V / RT) * (1/M_oxygen) * (1/1000)
Since we don't have the value of V (volume), we can cancel it out by dividing the equation by V:
C_oxygen = (P_oxygen / RT) * (1/M_oxygen) * (1/1000)
Now, let's substitute the given values:
C_oxygen = (2.1 * 10^4 N/m^2) / (8.314 J/mol·K * 300 K) * (1/32 g/mol) * (1/1000)
C_oxygen ≈ 8.42 * 10^-3 kg mol/m^3
Therefore, the molar concentration of oxygen is approximately 8.42 * 10^-3 kg mol/m^3, which corresponds to option (c).
Given:
Partial pressure of oxygen (P_oxygen) = 0.21 * (1 * 10^5 N/m^2) = 2.1 * 10^4 N/m^2
Partial pressure of nitrogen (P_nitrogen) = 0.79 * (1 * 10^5 N/m^2) = 7.9 * 10^4 N/m^2
Total pressure of the mixture (P_total) = 1 * 10^5 N/m^2
Temperature (T) = 300 K
The ideal gas law equation is given as:
PV = nRT
Where:
P = pressure
V = volume
n = moles of gas
R = ideal gas constant
T = temperature
We can rearrange the equation to solve for the moles of gas:
n = PV / RT
Now, let's calculate the moles of oxygen:
n_oxygen = P_oxygen * V / RT
Since we are looking for the molar concentration, we need to divide the moles of oxygen by the volume. The molar concentration is given in units of kg mol/m^3. To convert the molar concentration from mol/m^3 to kg mol/m^3, we need to divide by the molar mass of oxygen.
The molar mass of oxygen (M_oxygen) is approximately 32 g/mol.
To convert the molar concentration to kg mol/m^3, we use the following conversion factor:
1 mol/m^3 = (1/1000) kg mol/m^3
Therefore, the molar concentration of oxygen (C_oxygen) is given by:
C_oxygen = (n_oxygen / V) * (1/M_oxygen) * (1/1000)
Now, let's substitute the values into the equation:
C_oxygen = (P_oxygen * V / RT) * (1/M_oxygen) * (1/1000)
Since we don't have the value of V (volume), we can cancel it out by dividing the equation by V:
C_oxygen = (P_oxygen / RT) * (1/M_oxygen) * (1/1000)
Now, let's substitute the given values:
C_oxygen = (2.1 * 10^4 N/m^2) / (8.314 J/mol·K * 300 K) * (1/32 g/mol) * (1/1000)
C_oxygen ≈ 8.42 * 10^-3 kg mol/m^3
Therefore, the molar concentration of oxygen is approximately 8.42 * 10^-3 kg mol/m^3, which corresponds to option (c).
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A binary mixture of oxygen and nitrogen with partial pressures in the ratio 0.21 and 0.79 is contained in a vessel at 300 K. If the total pressure of the mixture is 1 * 105N/m2, find molar concentration of oxygena)8.42 * 10-1kg mol/m3b)8.42 * 10-2kg mol/m3c)8.42 * 10-3kg mol/m3d)8.42 * 10-4kg mol/m3Correct answer is option 'C'. Can you explain this answer?
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