easy, do it as a proportion.
(.4-x)/.4=23/37
solve for x.
(.4-x)/.4=23/37
solve for x.
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the ideal gas constant
T is the temperature of the gas
In this problem, the volume of the tank and the temperature remain constant. Thus, we can rewrite the equation as:
P1V1 = nRT1 and P2V2 = nRT2
where P1 and P2 are the initial and final pressures respectively, and V1 and V2 are the initial and final volumes respectively.
We want to find the mass (in kg) of oxygen that must be withdrawn from the tank, so we need to find the number of moles first.
Rearranging the equation, we can solve for n:
n = P1V1 / RT1
Now, we can use the given information to find n:
P1 = 37 atm
V1 = unknown (The volume is not given, but we can assume it's constant)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T1 = unknown (The temperature is not given, but we can assume it's constant)
Next, we need to find the final number of moles of oxygen using the second equation:
n = P2V2 / RT2
P2 = 23 atm
V2 = same as V1
R = same as before
T2 = same as T1
Now that we have both initial and final number of moles, we can calculate the moles of oxygen that must be withdrawn:
Δn = n1 - n2
Finally, using the molar mass of oxygen (O2), which is approximately 32 g/mol, we can find the mass of oxygen that must be withdrawn:
Mass = Δn * molar mass
Remember to convert the mass to kg if necessary by dividing it by 1000.