A cylinder with a movable piston contains 2.00g 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 chanded from 2.00L to 2.70L? The temperature was held constant.
The easy way to explain but the long way to work it is the following:
Use PV = nRT. You know V = 2.00L, you can assume any value for T, you know n = 0.500(2.00 g/molar mass He). Solve for P.
Then for the 2.7O L volume, use PV = nRT, substitute P from the first calculation, T is the same as in the first, and solve for n. Multiply by molar mass to obtain grams and subtract from the original 2.00 to find how much additional He was added.
To solve this problem, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the temperature is held constant, we can simplify the equation to: PV = constant.
First, let's calculate the initial number of moles of helium in the cylinder:
Given:
Initial volume (V1) = 2.00 L
Pressure (P1) = constant
Initial mass of helium (m1) = 2.00 g
Molar mass of helium (M) = 4.00 g/mol
Number of moles (n1) = m1 / M
n1 = 2.00 g / 4.00 g/mol
n1 = 0.50 mol
Now, using the relationship PV = constant, we can calculate the final number of moles of helium in the cylinder:
Final volume (V2) = 2.70 L
Number of moles (n2) = n1 = 0.50 mol
Using the equation PV = constant, we get:
(P1)(V1) = (P2)(V2)
(P2) = (P1)(V1) / (V2)
(P2) = (constant)(2.00 L) / (2.70 L)
(P2) = (constant)(0.74)
Since the pressure remains constant, (constant) = (P1) = (P2).
Therefore, (P2) = (P1) = (constant) = (0.74).
Now, let's calculate the final mass of helium in the cylinder:
Final mass of helium (m2) = (n2)(M)
m2 = (0.50 mol)(4.00 g/mol)
m2 = 2.00 g
To determine how many grams of helium were added, we can subtract the initial mass from the final mass:
Grams of helium added = m2 - m1
Grams of helium added = 2.00 g - 2.00 g
Grams of helium added = 0 g
Therefore, no grams of helium were added to the cylinder since the pressure and temperature remained constant.
To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
In this scenario, we are told that the pressure remains constant, so we can rewrite the ideal gas law as V1/n1 = V2/n2, where V1 and V2 are the initial and final volumes, and n1 and n2 are the initial and final number of moles of gas.
Since we are only concerned with the change in the number of moles of helium, we can simplify the equation to V1/n1 = V2/n2.
Now, we can calculate the number of moles of helium initially present in the cylinder. Given that the initial volume is 2.00L and the pressure is constant, we can assume the number of moles remains constant as well.
Therefore, V1/n1 = V2/n2 can be rewritten as 2.00L/n1 = 2.70L/(n1 + n), where n is the number of moles of helium added.
Now, we can solve for n, the number of moles of helium added. By cross-multiplying and rearranging the equation, we have:
2.70L × n1 = 2.00L × (n1 + n)
2.70n1 = 2.00n1 + 2.00n
0.70n1 = 2.00n
n = (0.70n1) / 2.00
Finally, substitute the given values into the equation. The initial mass of helium is 2.00g. Since the molar mass of helium is 4.00 g/mol, we can calculate the number of moles initially present:
n1 = (2.00g) / (4.00 g/mol)
n1 = 0.50 mol
Now, substitute the value of n1 into the equation to find n:
n = (0.70 × 0.50 mol) / 2.00
n = 0.35 mol
To find the mass of helium added, multiply the number of moles (0.35 mol) by the molar mass of helium (4.00 g/mol):
Mass = (0.35 mol) × (4.00 g/mol)
Mass = 1.40 g
Therefore, 1.40 grams of helium were added to the cylinder.