3Review example 5.7 and exercise 5.7. Suppose there are 2.00 moles of O2 at 1520 mmHg and 27.0 oC. Rearrange the ideal gas law and use it to determine the volume (L) of the gas. Then, find the density by rearranging the equation P(Mm) = dRT. Show all units.
I put he numbers in the formula V=(nR/P)T which equals 24.6 L then I put it in the formula P(mm)/RT=d I got 998.6 g/l is this right?
I don't agree with that.I would have carried the 24.6 out another place and rounded later. That part is ok and I obtained 24.63 L; however, if you use P(Mm)= dRT you never need to go through the volume calculation since it isn't in the final equation. You just plug in
1520(32) = d(0.08206)(300.15) and solve for d.
You you like to do with calculating V first, that is ok; in that case you use
grams = 2 mols x molar mass and divide that by V to get g/L.
To find the volume (V) using the ideal gas law, we rearrange the equation PV = nRT and solve for V.
Given:
P = 1520 mmHg
n = 2.00 moles
T = 27.0 °C (convert to Kelvin by adding 273.15)
First, convert the pressure to standard units of atm:
1 atm = 760 mmHg
1520 mmHg / 760 mmHg/atm = 2 atm
Now, plug in the given values into the rearranged equation V = (nRT)/P:
V = (2.00 moles x 0.0821 L·atm/mol·K x (27.0 + 273.15) K) / 2 atm
V = 2.07 L
Therefore, the volume (V) of the gas is approximately 2.07 L.
To find the density, we use the equation P(Mm) = dRT, where P is pressure, Mm is the molar mass, d is the density, R is the ideal gas constant, and T is the temperature. Rearrange the equation to solve for density (d):
d = P(Mm) / RT
Given:
P = 1520 mmHg (already converted to atm)
Mm (molar mass) of O2 = 32.00 g/mol (16.00 g/mol per oxygen atom)
R = 0.0821 L·atm/mol·K
T = 27.0 °C (convert to Kelvin by adding 273.15)
Now, plug in the values and calculate:
d = (2 atm) * (32.00 g/mol) / (0.0821 L·atm/mol·K * (27.0 + 273.15) K)
d ≈ 1.165 g/L
Therefore, the density (d) of the gas is approximately 1.165 g/L.