A 3.70 L sample of neon at 11.24 atm is added to a 10.0 L cylinder that contains argon. If the pressure in the cylinder is 5.24 atm after the neon is added, what was the original pressure (in atm) of argon in the cylinder?
To find the original pressure of argon in the cylinder, we can use the principle of the ideal gas law, which states:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas (in Kelvin)
In this case, we have two gases: neon and argon.
Let's assume that the original pressure of argon in the cylinder is P_1 and the final pressure of the mixture after neon is added is P_2.
For the neon gas:
P_neon = 11.24 atm
V_neon = 3.70 L
For the argon gas:
P_argon = P_1 (unknown)
V_argon = 10.0 L
Since both gases are at the same temperature, we can assume that the temperature is constant. Therefore, the number of moles of the two gases combined is constant.
We can combine the volumes of the two gases to get the final volume:
V_final = V_neon + V_argon
Now, we can use the ideal gas law to find the number of moles for each gas:
n_neon = (P_neon * V_neon) / (R * T)
n_argon = (P_argon * V_argon) / (R * T)
Since the number of moles is constant, we can equate them:
n_neon = n_argon
Now, let's substitute the equations for the number of moles into the ideal gas law equation:
(P_neon * V_neon) / (R * T) = (P_argon * V_argon) / (R * T)
Canceling out the constants R and T, and substituting the given values, we have:
(11.24 atm * 3.70 L) = (P_argon * 10.0 L)
Rearranging the equation to solve for P_argon, we get:
P_argon = (11.24 atm * 3.70 L) / 10.0 L
P_argon = 4.15 atm
Therefore, the original pressure of argon in the cylinder was 4.15 atm.