If a 1.59 mole sample of argon occupies 3.07 L at constant temperature and pressure then what volume will 4.35 moles of argon occupy at the same conditions?
To solve this problem, we can use the ideal gas law, which states that the product of pressure (P) and volume (V) is directly proportional to the number of moles of gas (n) and the gas constant (R), and inversely proportional to the temperature (T). Mathematically, the ideal gas law is expressed as:
PV = nRT
where:
P = pressure (in units of pressure)
V = volume (in units of volume)
n = number of moles of gas
R = gas constant (8.314 J/(mol∙K))
T = temperature (in units of Kelvin)
We can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
Let's use this equation to find the volume for 4.35 moles of argon.
Given:
n1 = 1.59 moles of argon
V1 = 3.07 L
n2 = 4.35 moles of argon
T = constant (since it is given that the temperature is constant)
P = constant (since it is given that the pressure is constant)
Using these values, we can set up a proportion to solve for V2:
(V1 / n1) = (V2 / n2)
Substituting the given values:
(3.07 L / 1.59 mol) = (V2 / 4.35 mol)
Now, we can solve for V2 by rearranging the equation:
V2 = (3.07 L / 1.59 mol) * 4.35 mol
V2 ≈ 8.39 L
Therefore, 4.35 moles of argon will occupy approximately 8.39 L at the same temperature and pressure conditions.