A cylinder with a movable piston contains 2.00 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 3.10 l ?
yes thank you!!!!
That answer is not right.
To determine the number of grams of helium that were added to the cylinder, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (constant)
V = volume
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (assumed constant)
In this case, the pressure is constant, so we can rewrite the equation as:
V1/n1 = V2/n2
Where:
V1 = initial volume (2.00 L)
n1 = initial number of moles (unknown)
V2 = final volume (3.10 L)
n2 = final number of moles (unknown)
First, we need to determine the initial number of moles of helium in the cylinder.
Rearranging the equation, we have:
n1 = (V1 * n2)/V2
Substituting the given values:
n1 = (2.00 L * n2) / 3.10 L
Next, we need to convert the number of moles to grams using the molar mass of helium.
The molar mass of helium (He) is approximately 4.00 g/mol.
Now, we can calculate the grams of helium added:
grams of helium added = (n2 - n1) * molar mass of helium
Let's solve the equation step by step:
1. Calculate n1:
n1 = (2.00 * n2) / 3.10
2. Calculate n2:
Since the number of moles remains the same (constant pressure), n2 = n1
3. Convert moles to grams:
grams of helium added = (n2 - n1) * molar mass of helium
By substituting n2 = n1 in the above equation, we get:
grams of helium added = (n1 - n1) * molar mass of helium
= 0 * molar mass of helium
= 0 grams
Therefore, no helium was added to the cylinder, as the pressure remained the same while adjusting the volume.
Typo in the first line, I suppose you mean 2 grams of Helium
P1 V1 = n1 R T1
P1 V2 = n2 R T1
so
V1/V2 = n1/n2
2/3.1 = 2/x
x = 3.1
x-2 = amount added = 1.1