Question

1. If a balloon holds 2000 L of helium at sea level, at 1 atm and 25 °C, what will its vol- ume be if it is allowed to rise to a height of 8 miles, where the pressure is 0.30 atm and the temperature is-70°C? 2. What is the mass of 40.0 L of NH3 at 1.50 atm and 45 °C?

Answer 1

Ans. #1. #Step 1: Calculate moles of He gas:

Given, Initial conditions at 25.0⁰C

Volume, V = 2000.0 L          ; Pressure, P = 1.0 atm         ; Temperature, T = 298.15 K

# Using Ideal gas equation:            PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in equation 1-

            1.00 atm x 2000.0 L = n x (0.0821 atm L mol^-1K^-1) x 298.15 K

            Or, 2000.0 atm L = n x 28.735 atm L mol^-1

            Or, n = 2000.0 atm L / (24.478115 atm L mol^-1)

            Hence, n = 81.70564 mol

Therefore, moles of He in the balloon = 81.70564 mol

# Step 2: Using “n”, calculate the new volume

            Pressure, P = 0.30 atm         ; Temperature, T = -70.0⁰C = 203.15 K

Putting the values in equation 1-

            0.30 atm x V = 81.70564 mol x (0.0821 atm L mol^-1K^-1) x 203.15 K

            Or, 0.30 atm x V = 1362.7369128886 atm L

            Or, V = 1362.7369128886 atm L / 0.30 atm = 4542.45 L

Hence, volume of balloon = 4542.45 L

#2. Pressure, P = 1.50 atm ; Temperature, T = 45.0⁰C = 318.15 K

1.50 atm x 40.0 L = n x (0.0821 atm L mol^-1K^-1) x 318.15 K

            Or, 60.0 atm L = n x 26.120115 atm L mol^-1

            Or, n = 60.0 atm L / (26.120115 atm L mol^-1)

            Hence, n = 2.2971 mol

Now,

            Mass of NH₃ = moles x Molar mass

                                    = 2.2971 mol x (17.03056 g/ mol)

                                    = 39.121 g