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 4.50 ? (The temperature was held constant.)
Express your answer numerically in grams.
no units again.
That's like going to the store and asking the price of a gallon of milk. The clerk tells you 2. What does that mean? Absolutely nothing. It could be 2 mils, 2 cents, 2 dollars, 2 euros, 2 lira, what?
To solve this problem, we need to use the ideal gas law equation: 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 problem, the pressure and temperature remain constant. Therefore, we can simplify the equation to:
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, respectively.
Given that V1 = 2.00 L and V2 = 4.50 L, we can rearrange the equation to solve for n2:
n2 = (V2/V1) * n1
Now, let's calculate:
n2 = (4.50 L / 2.00 L) * 2.00 mol = 4.50 mol
We know that 1 mole of helium has a molar mass of 4.00 g. So, to find out how many grams of helium were added to the cylinder, we multiply the number of moles by the molar mass:
grams of helium added = n2 * molar mass
= 4.50 mol * 4.00 g/mol
= 18.00 g
Therefore, 18.00 grams of helium were added to the cylinder.