0.600 L of Ar at 1.20atm and 227C is mixed with 0.200 L of O2 at 501 torr and 127 C in a 400ml flask at 27C. What is the pressure in the flask?
This was a confusing problem when first posted a few days ago and it's no less confusing to me now. However, I think what you must do is to use PV = nRT for the first gas and calculate n for Ar gas. Then use PV = nRT and calculate n for O2. Finally, add the moles and use PV = nRT again, this time using 27 C and 400 mL with the combined n.
To find the pressure inside the flask after mixing, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin for both gases:
For Argon (Ar):
T1 = 227°C + 273.15 = 500.15 K
For Oxygen (O2):
T2 = 127°C + 273.15 = 400.15 K
Now let's find the number of moles of each gas:
For Argon (Ar):
n1 = PV/RT
= (1.20 atm * 0.600 L) / (0.0821 L.atm/mol.K * 500.15 K)
= 0.01465 mol
For Oxygen (O2):
n2 = PV/RT
= (0.501 atm * 0.200 L) / (0.0821 L.atm/mol.K * 400.15 K)
= 0.01211 mol
Next, let's calculate the total number of moles:
ntotal = n1 + n2
= 0.01465 mol + 0.01211 mol
= 0.02676 mol
Now, we need to find the total volume of the gases after mixing. Since the gases are initially in separate containers, their volumes can be added:
Vtotal = V1 + V2
= 0.600 L + 0.200 L
= 0.800 L
However, we need to account for the expansion of the gases when they are mixed. The flask's volume is given as 400 mL (or 0.400 L). Therefore, the total volume of the mixed gases is limited to the volume of the flask.
Vtotal = 0.400 L
Finally, let's calculate the pressure inside the flask using the ideal gas law:
Ptotal = ntotalRT/Vtotal
= (0.02676 mol * 0.0821 L.atm/mol.K * 500.15 K) / 0.400 L
= 20.813 atm
Therefore, the pressure inside the flask after mixing is approximately 20.813 atm.