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A rigid tank contains 1.40 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 Determine the number of moles of gas that must be withdrawn from the tank to lower the pressure of the gas from 21.6 atm to 5.90 atm. Assume the volume of the tank and the temperature of the gas remain constant during this operation.
To solve this problem, we can use the ideal gas law equation: PV = nRT, where
P is the pressure of the gas,
V is the volume of the tank,
n is the number of moles of gas,
R is the ideal gas constant (0.0821 L·atm/(mol·K)),
and T is the temperature of the gas.
Since the volume of the tank and the temperature of the gas remain constant, we can rearrange the equation as follows:
(P1)(V) = (n)(R)(T)
(P2)(V) = (n - x)(R)(T)
We are given:
P1 = 21.6 atm (initial pressure)
P2 = 5.90 atm (final pressure)
n = 1.40 moles (initial number of moles)
Now, we can solve for x, the number of moles of gas that must be withdrawn from the tank:
(21.6 atm)(V) = (1.40 moles)(R)(T)
(5.90 atm)(V) = (1.40 moles - x)(R)(T)
We can divide the two equations to eliminate the volume (V):
(21.6 atm) / (5.90 atm) = (1.40 moles)(R)(T) / (1.40 moles - x)(R)(T)
Simplifying, we get:
3.661 = 1 / (1 - x / 1.40)
Next, we solve for x:
3.661 = 1 / (1 - x / 1.40)
3.661(1 - x / 1.40) = 1
3.661 - 3.661x / 1.40 = 1
3.661 - 3.661x = 1.40
-3.661x = 1.40 - 3.661
-3.661x = -2.261
x = -2.261 / -3.661
x ≈ 0.617 moles
Therefore, approximately 0.617 moles of gas must be withdrawn from the tank to lower the pressure from 21.6 atm to 5.90 atm.