A rigid tank contains 1.80 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to lower the pressure of the gas from 21.6 atm to 5.80 atm. Assume the volume of the tank and the temperature of the gas remain constant during this operation.
1.8 (1 - 5.80/21.6) = 1.32 moles
To determine the number of moles of gas that must be withdrawn from the tank, we can utilize the ideal gas law, which states:
PV = nRT
Where:
P = pressure of the gas
V = volume of the gas
n = number of moles of gas
R = ideal gas constant
T = temperature of the gas
Since the volume and temperature remain constant, we can simplify the equation to:
P1 * n1 = P2 * n2
Where:
P1 = initial pressure of the gas (21.6 atm)
n1 = initial number of moles of gas (1.80 moles)
P2 = final pressure of the gas (5.80 atm)
n2 = final number of moles of gas (unknown)
Rearranging the equation to solve for n2, we have:
n2 = (P1 * n1) / P2
Plugging in the values:
n2 = (21.6 atm * 1.80 moles) / 5.80 atm
Simplifying the expression:
n2 = 6.732 moles / 5.80 atm
n2 ≈ 1.16 moles
Therefore, approximately 1.16 moles of gas must be withdrawn from the tank to lower the pressure from 21.6 atm to 5.80 atm.