Answer to Question #256397 in General Chemistry for Gayle

Question

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 2.70 

L? (The temperature was held constant.)

Answer 1

Given;

Mass of Helium (m1) = 2.00g

Temperature remains same (T) is constant

Pressure remains (P) is constant

Volume (V1) = 2.00L

Volume (V2) = 2.70L

According to Ideal gas law,

PV = nRT(where R is constant)

$\frac{PV}{nt} =constant$

$\frac{P1V1}{n1T1}=\frac{P2V2}{n2T2}$ (P and T are constant)

$\frac{V1}{n1}=\frac{V2}{n2}$

$n=\frac{mass}{mw}$

Since molecular mass = constant

$\frac{V1}{mass1} =\frac{V2}{mass2}$

$mass2= V2×\frac{mass1}{V1}$

$= \frac{2.70L×2.00g}{2.00L}$

$mass2 = 2.70g$

$=mass2-mass1$

$=2.70g - 2.00g = 0.70g$

Therefore 0.70g of He was added to the cylinder.