Respuesta :
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
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:
[tex]\frac{V_{1} }{n_{1} } =\frac{V_{2} }{n_{2} } \\\frac{2}{0.5} =\frac{3.9}{n_{2}} \\n_{2}=0.975moles[/tex]
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