An apparatus consists of a 4 L flask containing nitrogen gas at 32 C 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
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To find the partial pressure of nitrogen after the gases mix, 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
R is the ideal gas constant
T is the temperature
First, we need to calculate the number of moles of nitrogen and argon in each flask.
For the 4 L flask containing nitrogen gas:
Using the ideal gas law equation, we can rearrange it to solve for n:
n = PV / RT
Plugging in the values:
P = 885 kPa
V = 4 L
R = 8.314 L·kPa/(mol·K)
T = 32 °C + 273.15 K = 305.15 K
n_nitrogen = (885 kPa * 4 L) / (8.314 L·kPa/(mol·K) * 305.15 K)
Next, we calculate the number of moles of argon in the 8 L flask:
Using the same equation:
P = 47.5 kPa
V = 8 L
T = 32 °C + 273.15 K = 305.15 K
n_argon = (47.5 kPa * 8 L) / (8.314 L·kPa/(mol·K) * 305.15 K)
After finding the number of moles of nitrogen and argon, we can add them together since they mix together:
n_total = n_nitrogen + n_argon
Now, we need to find the partial pressure of nitrogen after mixing. To do this, we divide the moles of nitrogen by the total number of moles and multiply by the total pressure:
P_nitrogen = (n_nitrogen / n_total) * (885 kPa + 47.5 kPa)
Finally, we can substitute the calculated values into the equation to find the partial pressure of nitrogen after mixing.