Calculate the mass of oxygen gas required to

occupy a volume of 6 L at a pressure of 20.9
kPa and a temperature of 37◦C.

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PV=nRT

n=PV/RT
n=mass/32

mass=32*PV/RT

t in kelvins, choose R units to match your pressure volume units.

To calculate the mass of oxygen gas required, 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, R is the ideal gas constant, and T is the temperature.

1. Convert the temperature to Kelvin:
To convert Celsius to Kelvin, add 273.15. So, 37°C + 273.15 = 310.15 K.

2. Convert the pressure to atmospheres (atm):
Since the ideal gas constant, R, is in units of L*atm/mol*K, we need to convert the pressure from kilopascals (kPa) to atmospheres (atm). Divide the pressure by 101.325 (1 atm = 101.325 kPa), so 20.9 kPa / 101.325 = 0.206 atm.

Now, we have the following values to use in the ideal gas law equation:
P = 0.206 atm
V = 6 L
T = 310.15 K

3. Rearrange the ideal gas law equation to find n:
PV = nRT
n = PV / RT

Plug in the values:
n = (0.206 atm * 6 L) / (0.0821 L*atm/mol*K * 310.15 K)
n ≈ 0.0045 mol

4. Calculate the mass of oxygen gas:
The molar mass of oxygen (O₂) is approximately 32 g/mol.

mass = number of moles * molar mass
mass = 0.0045 mol * 32 g/mol
mass ≈ 0.144 g

Therefore, the mass of oxygen gas required to occupy a volume of 6 L at a pressure of 20.9 kPa and a temperature of 37°C is approximately 0.144 grams.

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