A sample of O2 gas is placed in a sealed 1.00 L flask and pressure is measured as temperature is changed. The data are plotted and the slope of the resulting line is equal to 0.00300 atm/K. The line passes through the origin of the graph. There are ???? g of O2 gas in the flask.
Since PV = nRT,
P/T (your slope)= nR/V = 0.0300 atm/K
n (the number of moles) = (V/R)*(P/T)
= (1 liter/0.08206 liter atm/mole K)* 0.003 atm K = 0.03656 moles
Convert that to grams
Here is how you convert it to grams
(0.03656 moles of O2)*[(32.0g of O2)/(1 mole of O2)]= ____g of O2
To determine the mass of O2 gas in the flask, we can make use of the Ideal Gas Law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
We can rearrange the equation to solve for the number of moles (n):
n = PV / RT
Given that the pressure (P) is measured as the temperature (T) changes, and the data is plotted to show a straight line passing through the origin of the graph, this suggests that we can assume the relationship between pressure and temperature is directly proportional.
Therefore, we can write:
P = kT
Where:
k = proportionality constant
Since the slope of the resulting line is given as 0.00300 atm/K, we can substitute this value for k in the equation:
P = (0.00300 atm/K) * T
Now we substitute the expression for P into the rearranged Ideal Gas Law equation:
n = [(0.00300 atm/K) * T] * V / (0.0821 L·atm/(mol·K))
Given that the volume (V) is 1.00 L, and assuming the temperature (T) is given (or if not, it can be provided), we can now calculate the number of moles (n) of O2 gas in the flask.
Finally, to calculate the mass of O2 gas, we need to multiply the number of moles by the molar mass of O2 (32 g/mol).
Mass = n * Molar mass
Substitute the value for n and the molar mass of O2:
Mass = (n) * (32 g/mol)
Using this information, you can determine the mass of O2 gas in grams.