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 Determine the number of moles of gas that must be withdrawn from the tank to lower the pressure of the gas from 27.9 atm to 4.80 atm. Assume the volume of the tank and the temperature of the gas remain constant during this operation.
Hints* Boyle's Law
Using P=(RnT)/V, where T=1(arbitrary constant), and R = 8.314: Calculate volume at initial condition. Use this volume and final pressure to calculate n final. Subtract n final from n inital.
To solve this problem, we can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the volume and temperature are constant in this case, we can simplify the equation to:
P1 * n1 = P2 * n2
where P1 and n1 are the initial pressure and number of moles of gas, and P2 and n2 are the final pressure and number of moles of gas.
Given:
P1 = 27.9 atm
n1 = 1.80 moles
P2 = 4.80 atm
Let's solve for n2:
P1 * n1 = P2 * n2
n2 = (P1 * n1) / P2
Substituting the given values:
n2 = (27.9 atm * 1.80 moles) / 4.80 atm
n2 ≈ 10.4625 moles
Therefore, the number of moles of gas that must be withdrawn from the tank to lower the pressure from 27.9 atm to 4.80 atm is approximately 10.4625 moles.