Given that many laboratory gases are sold in steel cylinders with a volume of 43.8 L, What mass (in grams) of argon is inside a cylinder whose pressure is 17810kPa at 24∘C?

Question

Given that many laboratory gases are sold in steel cylinders with a volume of 43.8 L, What mass (in grams) of argon is inside a cylinder whose pressure is 17810kPa at 24∘C?

I've attempted this problem but I'm not getting the correct answer. Please explain if you can!

Answer 1

To calculate the mass of argon inside the cylinder, we can use the ideal gas law equation:

PV = nRT

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

First, let's convert the given values to the correct units:
Pressure: 17810 kPa * 1000 = 17810000 Pa
Volume: 43.8 L * 0.001 = 0.0438 m^3
Temperature: 24 °C + 273.15 = 297.15 K

Now, rearranging the equation to solve for the number of moles (n):

n = PV / RT

n = (17810000 Pa * 0.0438 m^3) / (8.314 J/(mol⋅K) * 297.15 K)

n ≈ 278.16 mol

The molar mass of argon (Ar) is approximately 39.948 g/mol. So, to find the mass in grams, we can multiply the number of moles by the molar mass:

Mass = n * molar mass

Mass = 278.16 mol * 39.948 g/mol

Mass ≈ 11114.77 g

Therefore, the mass of argon inside the cylinder is approximately 11114.77 grams.

Answer 2

To calculate the mass of argon inside the cylinder, you can use the ideal gas law equation:

PV = nRT

Where:
P = Pressure (in Pascal)
V = Volume (in cubic meters)
n = Number of moles
R = Universal gas constant (8.314 J/(mol⋅K))
T = Temperature (in Kelvin)

First, let's convert the given values to appropriate units:
Pressure (P): 17810 kPa (kiloPascals) = 17810 × 1000 = 17,810,000 Pa
Volume (V): 43.8 L (liters) = 43.8 × 0.001 = 0.0438 m³
Temperature (T): 24 °C (Celsius) = 24 + 273.15 = 297.15 K

Now, rearranging the ideal gas law equation to solve for the number of moles (n):

n = PV / (RT)

Substituting the values we converted earlier:

n = (17,810,000 Pa) × (0.0438 m³) / ((8.314 J/(mol⋅K)) × (297.15 K))

Calculating this expression will give us the number of moles of argon inside the cylinder.

n = 26.8503 mol

Next, we'll convert the number of moles to mass using the molar mass of argon, which is approximately 39.95 g/mol. We can use the formula:

mass = n × molar mass

mass = 26.8503 mol × 39.95 g/mol

Calculating this expression will give us the mass of argon inside the cylinder.

mass ≈ 1073.31 g

Therefore, the mass of argon inside the cylinder is approximately 1073.31 grams.

Answer 3

p = 1.67*10^4 kilo Pascals

V = 43.8 L
T = 273 + 24 = 297 K
R - 8.314 Liters kilopascals/deg K

PV = n R T

n = (1.67*10^4)(43.8)/ [ 8.314*297 ]

n = 296 mols

296 * 40 grams/mol = 11840 grams
or 11.84 kg