1.9 moles of Ar and 2.5 moles of Ne are added to a 10.0 L container at 20.0oC. What will the internal pressure (atm) be?
Use PV = nRT with n = 1.9+2.5 = ?
10.58 atm
To find the internal pressure of the container, we can use the ideal gas law equation, which is stated as:
PV = nRT
Where:
P is the pressure in atm
V is the volume of the container in liters
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin
First, we need to convert the temperature from Celsius to Kelvin. The conversion is given by:
T(K) = T(°C) + 273.15
So, T(K) = 20.0°C + 273.15 = 293.15 K
Now, we have the following information:
- Volume (V) = 10.0 L
- Moles of Argon (n) = 1.9 mol
- Moles of Neon (n) = 2.5 mol
- Temperature (T) = 293.15 K
Since we are dealing with a mixture of gases, we need to calculate the total moles (n_total) before we can calculate the pressure.
n_total = n_Ar + n_Ne
n_total = 1.9 + 2.5 = 4.4 mol
Now, we can plug these values into the ideal gas law equation:
PV = nRT
P * 10.0 = 4.4 * 0.0821 * 293.15
10P = 4.4 * 24.0664
P = (4.4 * 24.0664) / 10
P = 10.5912
Therefore, the internal pressure of the container will be approximately 10.6 atm.