A cylinder with a movable piston contains 2.00 of helium, , 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 to 3.70 ? (The temperature was held constant.)
Pressure and temp the same?
You can do this by proportion then
I don't know what the 2.00 of helium means, but
gramsHeadded/2.00=1.70/2.00
yes
To find out how many grams of helium were added to the cylinder, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant
T = temperature of the gas
In this case, the pressure and temperature are constant. Therefore, the equation simplifies to:
V1/n1 = V2/n2
Where:
V1 = initial volume
n1 = initial number of moles
V2 = final volume
n2 = final number of moles
Given:
V1 = 2.00 L
V2 = 3.70 L
n1 = initial number of moles (unknown)
n2 = final number of moles (unknown)
Since the pressure is constant, we can write:
V1/n1 = V2/n2
To find n1, we can rearrange the equation:
n1 = n2 * V1 / V2
Now we can substitute the known values into the equation:
n1 = n2 * 2.00 L / 3.70 L
Since the number of moles remains the same, we can set n2 = n1:
n1 = n1 * 2.00 L / 3.70 L
Divide both sides of the equation by n1:
1 = 2.00 L / 3.70 L
Now we can solve for n1:
n1 = 1 * (3.70 L / 2.00 L)
n1 = 1.85
So, the initial number of moles (n1) is 1.85.
Since the question asks for the mass of helium added, we need to calculate the mass using the molar mass of helium. The molar mass of helium is approximately 4.00 g/mol.
To find the mass, we can use the equation:
mass = n * molar mass
Substituting the known values:
mass = 1.85 mol * 4.00 g/mol
mass = 7.40 g
Therefore, approximately 7.40 grams of helium were added to the cylinder.