An apparatus consists of a 4 L flask containing nitrogen gas at 32C and 885 kPa, joined by a valve to a 8 L flask containing argon gas at 32 C and 47.5 kPa. The valve is opened and
the gases mix. What is the partial pressure of nitrogen after mixing?
Answer in units of kPa
Effectively all you are doing is letting the N2 gas in the 4L container expand to a total of 12 L so the pressure of the N2 gas will be
885 x 4/12 = ? kPa.
no
To find the partial pressure of nitrogen after the gases mix, we can make use of the ideal gas law, which states that the pressure of a gas is proportional to its number of moles and temperature.
First, let's calculate the number of moles of nitrogen and argon in the initial state.
To find the number of moles, we can use the formula:
n = PV / RT
Where:
P = pressure (in pascals)
V = volume (in cubic meters)
R = ideal gas constant (8.314 J/(mol⋅K))
T = temperature (in Kelvin)
n = number of moles
Let's convert the given values into the appropriate units:
For nitrogen:
Pressure (P) = 885 kPa = 885,000 Pa
Volume (V) = 4 L = 0.004 m³
Temperature (T) = 32°C = 32 + 273.15 K
For argon:
Pressure (P) = 47.5 kPa = 47,500 Pa
Volume (V) = 8 L = 0.008 m³
Temperature (T) = 32°C = 32 + 273.15 K
Now, we can calculate the number of moles for each gas:
For nitrogen:
n₁ = (P₁ * V₁) / (R * T₁)
For argon:
n₂ = (P₂ * V₂) / (R * T₂)
Substituting the given values:
n₁ = (885,000 * 0.004) / (8.314 * (32 + 273.15))
n₁ ≈ 0.0556 mol
n₂ = (47,500 * 0.008) / (8.314 * (32 + 273.15))
n₂ ≈ 0.0895 mol
After the gases mix, their volumes add up:
V_total = V₁ + V₂
V_total = 4 L + 8 L
V_total = 12 L
Since the total number of moles remains constant, the number of moles of nitrogen in the final state will be:
n_final = n₁
Finally, we can calculate the partial pressure of nitrogen:
P_final = (n_final * R * T_total) / V_total
Substituting the values:
P_final = (0.0556 * 8.314 * (32 + 273.15)) / 12
P_final ≈ 120.8 kPa
Therefore, the partial pressure of nitrogen after mixing is approximately 120.8 kPa.