Question

A cylinder with a movable piston contains 2.00 gg of helium, HeHe, 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 LL to 3.90 LL ? (The temperature was held constant.)

Answer 1

Answer:

The grams of helium added is 1.9 g

Explanation:

According to the ideal gas expression and due pressure, volume and temperature are constant, we have that V/n = constant

The number of moles is:

$\frac{V_{1} }{n_{1} } =\frac{V_{2} }{n_{2} } \\\frac{2}{0.5} =\frac{3.9}{n_{2}} \\n_{2}=0.975moles$

The moles of He added is:

nHe = 0.975 - 0.5 = 0.475 moles

The mass of He added is

mHe = 0.475 * 4 = 1.9 g

Answer 2

Answer: 1.8 g

Explanation:

We start first, by calculating the amount of Helium

n = m/M

m = mass of Helium

M = molar mass if Helium

n = 2/4 = 0.5 moles

proceeding further, we use ideal gas law. PV = nRT

Then we have

P1V1/n1T1 = P2V2/n2T2

So that,

n2 = n1T1P2V2/P1V1T2

From the question, we know that, P1 = P2, and T1 = T2. So that,

n2 = n1v2/v1

n2 = (0.5 * 3.9) / 2

n2 = 1.95/2

n2 = 0.975 moles. With this, we can determine the mass, m2 of Helium

n = m/M

m = n * M

m = 0.975 * 3.9

m = 3.8

The difference between both masses are 3.8 - 2 = 1.8 g

Thus, 1.8 g of Helium was added to the cylinder