Determine the pressure in atm for a 10 g of CO2 sample contained in a sealed 10 L tank at 50degrees C?
Did anyone get 26.5 atm?
p =m•R•T/μ•V = =10•10^-3•8.31•323/44•10•10^-3 = 61 Pa = =6.02•10^4 atm.
I get 61. But after that then what?
I have something completely different.
V must go in in L.
P = nRT/V
n = mol = g/molar mass = 10/44.
P 10*0.o08206*(273+50)/10 = 0.60 atm.
Of course,
1 Pa = 9.869•10^-6 atm,
61 Pa = 6.02•10 ^ - 4 atm.
To determine the pressure in atm for a sample of CO2, you can use the Ideal Gas Law equation:
PV = nRT
where:
P is the pressure in atm
V is the volume in liters
n is the number of moles
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin
First, let's convert the given values to the correct units:
Mass of CO2 = 10 g
Volume of tank = 10 L
Temperature = 50 degrees C = 323 K (K = °C + 273.15)
Next, we need to find the number of moles of CO2. To do this, we use the molar mass of CO2, which is 44.01 g/mol.
Moles of CO2 = Mass of CO2 / Molar mass of CO2
Moles of CO2 = 10 g / 44.01 g/mol
Now, let's substitute the values into the Ideal Gas Law equation:
PV = nRT
P * 10 L = (10 g / 44.01 g/mol) * (0.0821 L·atm/(mol·K)) * 323 K
Simplifying the equation gives:
P = (10 g / 44.01 g/mol) * (0.0821 L·atm/(mol·K)) * 323 K / 10 L
Calculating this expression will give you the pressure in atm for the given sample.
Now, let's do the calculation:
P = (10 g / 44.01 g/mol) * (0.0821 L·atm/(mol·K)) * 323 K / 10 L
P ≈ 26.5 atm
Therefore, the pressure in atm for the given CO2 sample is approximately 26.5 atm.
If someone else obtained the same result of 26.5 atm, then their calculation is correct.