A gaseous mixture in a 25.00 L container is made of 16.0 g N2 and 14.0 g Ar and
has a total pressure of 1.00 atm.
1.Calculate the partial pressure of the N2 in the mixture. the ans is 0.620 atm
2.Calculate the temperature of the gas mixture.
can someone explain to me how to do the second part please?
use PV=nRT where n is the total number of moles.
Use V=25.00 L
P =1.00 atm
you needed to calculate n for the first part.
Make sure that use the correct value for R based on the units.
To calculate the temperature of the gas mixture, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas mixture
V is the volume of the container
n is the number of moles of gas
R is the ideal gas constant
T is the temperature of the gas in Kelvin
First, let's find the number of moles of N2 and Ar in the mixture.
To find the number of moles, we can use the formula:
n = m/M
Where:
n is the number of moles
m is the mass of the substance
M is the molar mass of the substance
For N2:
m = 16.0 g
M = 28.0134 g/mol (molar mass of N2)
n(N2) = 16.0 g / 28.0134 g/mol
n(N2) = 0.5712 mol
For Ar:
m = 14.0 g
M = 39.948 g/mol (molar mass of Ar)
n(Ar) = 14.0 g / 39.948 g/mol
n(Ar) = 0.3502 mol
Now, we can calculate the total number of moles of gas in the mixture:
n(total) = n(N2) + n(Ar)
n(total) = 0.5712 mol + 0.3502 mol
n(total) = 0.9214 mol
Next, let's rearrange the ideal gas law equation to solve for temperature:
T = PV / (nR)
Since we are given the pressure of the gas mixture (P = 1.00 atm), the volume of the container (V = 25.00 L), and the total number of moles of gas (n(total) = 0.9214 mol), we can substitute these values into the equation and solve for T.
T = (1.00 atm) * (25.00 L) / (0.9214 mol * R)
The value of the ideal gas constant (R) is 0.0821 L·atm/(mol·K).
Substituting the value of R, we get:
T = (1.00 atm) * (25.00 L) / (0.9214 mol * 0.0821 L·atm/(mol·K))
Simplifying the expression:
T = 25.00 L / (0.9214 mol * 0.0821 L·atm/(mol·K))
T ≈ 307.4 K
Therefore, the temperature of the gas mixture is approximately 307.4 K.