Question

EXAMPLE 6-11 Applying the Ideal Gas Equation to a Mixture of Gases What is the pressure, in bar, exerted by a mixture of 1.0 g H2 and 5.00 g He when the mixture is confined to a volume of 5.0 L at 20 °C? Analyze For fixed T and V, the total pressure of a mixture of gases is determined by the total number of moles of gas: Ptot = ntotRT/V. Solve ntot = (1.0g Hz x 1 mol Hy) - 65 X207032) + (5.00 g He x 1 mol He ) 4.003 g He = 0.50 mol H2 + 1.25 mol He = 1.75 mol gas 1.75 mol x 0.0831 bar L mol-1 K-1 X (20 + 273) K Ptot = - -= 8.5 bar 5.0 L Assess It is also possible to solve this problem by starting from equation (6.12). Because 1 mol of ideal gas occu- pies 22.7 L at 273 K and 1 bar, the pressure exerted by 1.75 mol of gas in a 5.0 L vessel at 293 K is (1.75 mol/1.00 mol) x (293 K/273 K) X (22.7 L/5.0 L) X 1.0 bar = 8.5 bar. PRACTICE EXAMPLE A: What will be the total gas pressure, in bar, if 12.5 g Ne is added to the mixture of gases described in Example 6-11 and the temperature is then raised to 55 °C? [Hint: What is the new number of moles of gas? What effect does raising the temperature have on the pressure of a gas at constant volume?]

Answer 1

Molar mass of neon = 20.1797 g/mol Calculate the moles of neon as follows: 1 mol Ne Moles of neon = 12.5 g Nex = 0.62 mol gas 20.1797 g Ne) Then not = 1.75+0.62 = 2.37 mol gas P tot Now calculate the total pressure as follows: 2.37 mol < 0.0831 bar L mol- K-4x(55+ 273) K 5L p_2.37 0.0831~(55+ 273) 5 Prot = 12.92 bar P =- tot Please give good rating.