Question

If 0.750 L of argon at 1.50 atm and177°C and 0.235 L of sulfur dioxide at 95.0 kPa and 63.0°Care added to a 1.00-L flask and the flask's temperature is adjustedto 25.0°C, what is the resulting pressure in the flask?

Answer 1

Moles of Argon, n = P V / RT
                            = 1.50 atm * 0.750 L / (0.0821 L-atm/mol/K *450K)

                             =0.03045 moles

Moles of SO2, n = 95.0 kPa * 0.235 L / (8.314 kPa-L/mol/K * 336 K)

                          =0.007991 moles

So total moles of gas = 0.03045 + 0.007991 = 0.03844moles

So pressure of the gas mixture, P = nRT / V

                                                   =0.03844 * 0.0821 L-atm/mol/K * 298 K /   1.00L

                                                   =0.9404 atm