Question

A 5.00-L flask contains nitrogen gas at 25°C and 1.00 atm pressure. What is the final pressure in the flask if an additional 2.00 g of N2 gas is added to the flask and the flask cooled to -55°C?

a.) 0.987 atm

b.) 1.35 atm

c.) 1.84 atm

d.) 0.255 atm

Answer 1

Ans. # Calculate initial moles of N2 using ideal gas equation-

            # Now, 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 0-1

            1.00 atm x 5.00 L = n x (0.0821 atm L mol^-1K^-1) x 298.15 K

            Or, 5.00 atm L = n x 24.478115 atm L mol^-1

            Or, n = 5.00 atm L / 24.478115 atm L mol^-1

            Hence, n = 0.003354 mol

# Mole of N2 added to vessel = Mass of N2 / Molar mass

                                    = 2.0 g / (28.01348 g / mol)

                                    = 0.071394 mol

# Total moles of N2 in vessel = Initial moles + Moles added

                                    = 0.003354 mol + 0.071394 mol

                                    = 0.074748 mol

# Calculate the pressure using equation 1-

Putting the new set of values (at -55.0⁰C) in equation 1-

            P x 5.00 L = 0.074748 x (0.0821 atm L mol^-1K^-1) x 218.15 K

            Or, P = 1.33875 atm L / 5.00 L

            Hence, P = 0.2677 atm

Hence, pressure in vessel, P = 0.2677 atm