Acylinder 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 3.70 l ? (the temperature was held constant.)

Hey There!

Ideal gas equation: PV = nRT where P is pressure, V is volume, n is moles of gas, R is molar gas constant and T is temperature.

Since P, R and T are constant, V/n = constant

V1 = 2.00 L, V2 = 3.30 L

n1 = (2.00 g)/(4.00 g/mol) = 0.5 mol, n2 = ?

V1/n1 = V2/n2

2.00/0.5 = 3.30/n2

n2 = 0.825 mol

Moles of He added = n2 - n1

= 0.825 - 0.5 = 0.325 mol

Mass of He added = moles x atomic mass

= 0.325 * 4.00 = 1.30 g of He

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .

a). in this case, dh> 0 and sd> 0. if t becomes larger, the term "-tds" will become more negative. at some point, it is possible that -dts will be more negative than the dh is positive, and the reaction will become spontaneous. this is probably the correct answer, but you should look at all of the other possibilities and make sure of that.