contestada

If 4.73 g Ar are added to 3.22 atm He in a 2.00 L cylinder at 27.0 °C. what is the total pressure of the resulting gaseous mixture?

Respuesta :

dsdrajlin

Answer:

4.67 atm

Explanation:

Step 1: Given data

  • Mass of Ar (mAr): 4.73 g
  • Partial pressure of He (pHe): 3.22 atm
  • Volume of the container (V): 2.00 L
  • Temperature (T): 27.0 °C

Step 2: Convert "T" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 27.0 °C + 273.15 = 300.2 K

Step 3: Calculate the moles of Ar

The molar mass of Ar is 39.94 g/mol.

4.73 g × 1 mol/39.94 g = 0.118 mol

Step 4: Calculate the partial pressure of Ar

We will use the ideal gas equation.

pAr × V = nAr × R × T

pAr = nAr × R × T / V

pAr = 0.118 mol × (0.0821 atm.L/mol.K) × 300.2 K / 2.00 L

pAr = 1.45 atm

Step 5: Determine the total pressure of the gaseous mixture

The total pressure of the gaseous mixture is equal to the sum of the partial pressures of the individual gases.

P = pAr + pHe = 1.45 atm + 3.22 atm = 4.67 atm