A 7.30-L container holds a mixture of two gases at 45 °C. The partial pressures of gas A and gas B, respectively, are 0.342 atm and 0.527 atm. If 0.250 mol a third gas is added with no change in volume or temperature, what will the total pressure become?
Use PV = nRT and solve for P using n = total mols of all three gases.
To find the total pressure after adding the third gas, we need to consider the partial pressures of all three gases.
Given:
Volume (V) = 7.30 L
Temperature (T) = 45 °C = 318.15 K
Partial pressure of gas A (PA) = 0.342 atm
Partial pressure of gas B (PB) = 0.527 atm
Moles of the third gas (n3) = 0.250 mol
First, let's convert the temperature to Kelvin, as it is the standard unit for temperature in gas calculations.
T = 45 °C + 273.15 = 318.15 K
Next, we need to calculate the moles of the first two gases (gas A and gas B). To find the moles, we can use the ideal gas law equation:
PV = nRT
Rearranging the equation:
n = PV / RT
where:
P is the pressure
V is the volume
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
Moles of gas A (nA):
nA = PA * V / RT
nA = (0.342 atm) * (7.30 L) / (0.0821 L·atm/(mol·K) * 318.15 K)
nA ≈ 13.02 mol
Moles of gas B (nB):
nB = PB * V / RT
nB = (0.527 atm) * (7.30 L) / (0.0821 L·atm/(mol·K) * 318.15 K)
nB ≈ 19.94 mol
Now, we have the moles of gas A (nA), gas B (nB), and the moles of the third gas (n3). The total moles of gas can be calculated by adding these values:
Total moles (nTotal) = nA + nB + n3
nTotal = 13.02 mol + 19.94 mol + 0.250 mol
nTotal ≈ 33.21 mol
Finally, to find the total pressure (PTotal), we can use the ideal gas law again:
PTotal = nTotal * RT / V
PTotal = (33.21 mol) * (0.0821 L·atm/(mol·K)) * 318.15 K / 7.30 L
PTotal ≈ 11.47 atm
Therefore, the total pressure after adding the third gas will be approximately 11.47 atm.