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 0.410 atm and 0.620 atm. If 0.150 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
Total Pressure= PressureA + PressureB.....
when you add .150 mol you have to find pressure using PV=nrT sooooo.... P(7.30)=(.150)(.08206)(318) for P=.536 where n is moles, r is a constant and T is in kelvin... now add up all the Pressures
Ptotal= .410+.620+.536= 1.566
To find the total pressure after adding the third gas, we need to use the ideal gas law equation:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin
First, we need to determine the initial total pressure of the mixture before adding the third gas. The partial pressures of gas A and gas B can be added to find the initial total pressure.
P(initial) = P(A) + P(B)
P(initial) = 0.410 atm + 0.620 atm
P(initial) = 1.03 atm
Next, we need to convert the initial temperature from Celsius to Kelvin.
T(K) = T(°C) + 273.15
T(K) = 45 °C + 273.15
T(K) = 318.15 K
Now, we can use the ideal gas law to find the number of moles of the initial mixture.
PV = nRT
Rearranging the equation to solve for n, we have:
n = PV / RT
For the initial mixture:
n(initial) = P(initial) * V / (R * T)
n(initial) = (1.03 atm) * (7.30 L) / (0.0821 L*atm/mol*K * 318.15 K)
Calculating n(initial) gives us:
n(initial) = 0.283 mol
Next, we need to add the moles of the third gas to the initial number of moles:
n(final) = n(initial) + 0.150 mol
n(final) = 0.283 mol + 0.150 mol
n(final) = 0.433 mol
Finally, we can calculate the final pressure using the ideal gas law again:
P(final) = n(final) * R * T / V
P(final) = (0.433 mol) * (0.0821 L*atm/mol*K) * (318.15 K) / (7.30 L)
Calculating P(final) gives us:
P(final) = 1.88 atm
Therefore, the total pressure after adding the third gas will be approximately 1.88 atm.