How many grams of oxygen gas are there in 2.3L tank at 7.5atm and 24 degree celsius?
PV=nRT
n=PV/RT
but n= mass/molmass
so
mass=molmasso2*P*V/RT
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To find the number of grams of oxygen gas in the tank, you will first need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atmospheres (atm)
V = volume in liters (L)
n = number of moles of gas
R = the ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (K)
First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 24°C + 273.15 = 297.15K
Now, rearrange the ideal gas law equation to solve for n (number of moles of gas):
n = PV / RT
Next, substitute the given values into the equation:
P = 7.5 atm
V = 2.3 L
R = 0.0821 L·atm/mol·K
T = 297.15K
n = (7.5 atm * 2.3 L) / (0.0821 L·atm/mol·K * 297.15K)
Calculate the value of n:
n ≈ 0.6896 mol (rounded to four decimal places)
To find the mass of oxygen gas (O2), you need to know the molar mass of oxygen, which is approximately 32 g/mol.
Finally, multiply the number of moles (n) by the molar mass of oxygen (32 g/mol):
Mass = n * Molar mass
Mass = 0.6896 mol * 32 g/mol
Mass ≈ 22.0691 grams (rounded to four decimal places)
Therefore, there are approximately 22.0691 grams of oxygen gas in the 2.3L tank at 7.5 atm and 24 degrees Celsius.