How would I find the volume occupied by 3 moles of N2 at zero degrees and 101.325 kPa?

Use the equation,

PV = nRT

n = 3mol
T = 273K
P = 101.325kPa = 1atm
R = .08206L•atm/K•mol

Solve for V

To clarify David's response, P must be in atm (1 atm) if R is to be used as 0.08205.

A closed container is filled with oxygen. The pressure in the container is 185kPa . What is the pressure in millimeters of mercury?

Well, picture this: 3 moles of N2 having a little pool party inside a box at zero degrees and 101.325 kPa. Now, to find the volume occupied by these lively moles, you'll need to use the ideal gas law, which states PV = nRT. You already have the information about pressure (P), temperature (T), and moles (n). Now all you need is the gas constant (R). Once you got that, rearrange the equation to solve for volume (V) and you'll have the volume these moles are splashing around in. Enjoy the pool party!

To find the volume occupied by 3 moles of N2 at zero degrees (273.15 K) and 101.325 kPa, you would need to use the Ideal Gas Law equation, which is:

PV = nRT

Where:
- P is the pressure (in Pascals)
- V is the volume (in cubic meters)
- n is the number of moles
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature (in Kelvin)

First, convert the pressure to Pascals. Since 1 kPa = 1000 Pa, multiply 101.325 kPa by 1000 to get 101,325 Pa.

Next, plug in the known values into the equation:

(101,325 Pa) * V = (3 moles) * (8.314 J/(mol·K)) * (273.15 K)

Now we can solve for V. Rearrange the equation to isolate V:

V = (3 moles * 8.314 J/(mol·K) * 273.15 K) / (101,325 Pa)

Calculating the right side of the equation:

V ≈ 0.065 m³

Therefore, the volume occupied by 3 moles of N2 at zero degrees and 101.325 kPa is approximately 0.065 cubic meters.