The total pressure of a gas mixture containing 1.00g of H2 and 8.00g of Ar in a 3.0L container at 27°C is calculated using the ideal gas law, yielding a total pressure of approximately 5.77 atm.
To calculate the total pressure of a gas mixture that contains 1.00g of H2 and 8.00g of Ar in a 3.0L container at 27°C, we employ the ideal gas law, which is PV=nRT. For a mixture, the total pressure is the sum of the partial pressures of the individual gases. First, we need to find the number of moles (n) of each gas using their molar masses (H2 = 2.016 g/mol and Ar = 39.948 g/mol). Then, we convert the temperature to Kelvin by adding 273.15 to the Celsius temperature.
Number of moles of H2 = 1.00 g / 2.016 g/mol = 0.496 moles
Number of moles of Ar = 8.00 g / 39.948 g/mol = 0.200 moles
Total moles (n) = 0.496 moles H2 + 0.200 moles Ar = 0.696 moles
Temperature (T) in Kelvin = 27°C + 273.15 = 300.15 K
The ideal gas constant (R) is 0.0821 atm·L/mol·K
Now we can use the ideal gas law to find the total pressure (P).
PV = nRT
P = nRT / V
P = (0.696 moles) x (0.0821 atm·L/mol·K) x (300.15 K) / (3.0 L)
P = 5.77 atm (rounded to two decimal places)
The total pressure in atmospheres of the gas mixture is therefore 5.77 atm.