Answer:

Answer on Question #66231, Chemistry,General ChemistryPart C

A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 4.10 L ? (The temperature was held constant.)Solution:

Data:m1(He) = 2.00 gt1= 250CV1= 2.00 LV2= 4.10 LP1= P2T1= T2

Solution:First of all, we can calculate the amount of helium, using equation:n1= m1(He)M(He), where M(He) is molar mass of helium, according to Periodic table M(He) = 4.00 g/mol.Thus:n1= 2.004.00= 0.5 (moles)According to general relationship,known as theideal gas equation:PV=nRTWe can calculate the number of moles,when volume increases:

Where:n2 = n1∙T2∙P2∙V2P1∙V1∙T2Using given conditions (P1= P2, T1= T2) and values:n2 = n1∙V2V1= 0.5∙4.102.00= 1.025 (moles)After that we can determine the mass m2of helium gas:m2(He) = n2∙M(He) = 1.025∙4.00 = 4.10 (g)

Hope it helps.