Aballoon containing 0.500 mol at 0.00°c and 65.0kpa pressure is expanded by adding more argon. how many moles of argon are added to bring the sample to a final volume of 60.0 l at 30.0°c and 45.0kpa

n = 0.572 moles of Argon added to reach a volumen of 60.0 L

Explanation:

∴ n1 = 0.500 mol

∴ T1 = 0.0°C = 273.15 K

∴ P1 = 65.0 KPa * ( 0.009869 atm / KPa ) = 0.6415 atm

∴ R = 0.082 atm.L/K.mol

⇒ V1 = RT1n1 / P1 = 17.46 L

∴ P2 = 45.0 KPa = 0.444 atm

∴ V2 = 60.0 L

∴ T2 = 30.0 °C = 303.15 K

⇒ n2 = ((0.444 atm) * (60.0 L)) / ((0.082 atm.L/K.mol)*(303.15 K))

⇒ n2 = 1.072 mol

moles of Argon added:

n = n2 - n1 = 1.072 - 0.500 = 0.572 mol added

screening

explanation: