A rigid tank contains 1.20 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 Determine the number of moles of gas that must be withdrawn from the tank to lower the pressure of the gas from 28.5 atm to 5.30 atm. Assume the volume of the tank and the temperature of the gas remain constant during this operation.
__mol
use fomular P*v=n*R*T
cause temperature and volumn not change:
n(initial)=(P(initial)*V)/(R*T)
n(final)=(P(final)*V)/(R*T)
we want the final n, therefore:
n(final)=P(final)*n(initial)/P(initial)
=5.3*1.2/28.5=0.223
To determine the number of moles of gas that must be withdrawn from the tank, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure of the gas
V = volume of the tank (constant)
n = number of moles of gas
R = ideal gas constant
T = temperature of the gas (constant)
Since the volume and temperature of the gas remain constant, we can write the equation as:
P1 * V = n1 * R * T
P2 * V = n2 * R * T
where:
P1 = initial pressure of the gas (28.5 atm)
P2 = final pressure of the gas (5.30 atm)
V = volume of the tank (constant)
n1 = initial number of moles of gas (1.20 moles)
n2 = final number of moles of gas (unknown)
First, we need to find the value of R. The ideal gas constant (R) is typically given as 0.0821 L·atm/mol·K.
Now, we can rearrange the equation to solve for n2:
P1 * V / (R * T) = n1
P2 * V / (R * T) = n2
Substituting the given values:
(28.5 atm) * V / (0.0821 L·atm/mol·K * T) = 1.20 moles
(5.30 atm) * V / (0.0821 L·atm/mol·K * T) = n2
Now we can solve for n2:
n2 = (5.30 atm) * V / (0.0821 L·atm/mol·K * T)
Let's assume the volume of the tank is 1 liter (V = 1 L) and the temperature is 298 K (T = 298 K). Plugging in these values:
n2 = (5.30 atm) * (1 L) / (0.0821 L·atm/mol·K * 298 K)
n2 = 0.215 moles
Therefore, the number of moles of gas that must be withdrawn from the tank to lower the pressure from 28.5 atm to 5.30 atm is approximately 0.215 moles.