A gas storage container in an ordinary chemical laboratory measures 4.9 cm wide and 20 cm high. The label reads: N2 gas with pressure of 16.7 atm. If the cylinder is opened in the gas allowed to escape into a large, empty plastic bag, what will be the final volume of nitrogen gas, including what’s collected in the plastic bag and what’s left over in the cylinder write your answer in liters and round to two significant digits.
To find the final volume of nitrogen gas, we need to calculate the sum of the volume of gas collected in the plastic bag and the remaining volume in the cylinder.
First, we need to convert the dimensions of the cylinder to volume. The volume of a cylinder can be calculated using the formula: V = πr^2h, where V is the volume, r is the radius, and h is the height.
Given that the width of the cylinder is 4.9 cm, we can divide it by 2 to get the radius: r = 4.9 cm / 2 = 2.45 cm = 0.0245 m.
The height of the cylinder is 20 cm = 0.20 m.
Calculating the volume of the cylinder: V_cylinder = π(0.0245 m)^2 * 0.20 m = 9.47 * 10^(-4) m^3.
Next, we need to convert the pressure from atm to Pascal (Pa). 1 atm is approximately equal to 101325 Pa.
Pressure of nitrogen gas = 16.7 atm * 101325 Pa/atm = 1.689355 * 10^6 Pa.
To find the number of moles of nitrogen gas, 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 gas constant, and T is the temperature (which is assumed constant in this case).
Since we are only interested in the volume and not the number of moles, we can rearrange the equation to solve for V:
V = (nRT) / P.
Since the temperature and the gas constant are constant, we can rewrite the equation as:
V ∝ n.
This means that the volume is directly proportional to the number of moles of gas.
If we assume that the temperature and the number of moles of the gas do not change when the cylinder is opened and the gas is allowed to escape, then the final volume of nitrogen gas will be the sum of the volume in the plastic bag and the volume remaining in the cylinder.
Now, let's assume that when the gas is allowed to escape, the volume in the plastic bag increases while the volume in the cylinder decreases. We can represent the final volume of nitrogen gas as:
V_final = V_plastic bag + V_cylinder.
Since the volume is directly proportional to the number of moles of gas, we can assume that the final number of moles of nitrogen gas will be the same as the initial number of moles.
V_final ∝ n.
Therefore, V_final is directly proportional to the initial volume.
Similarly, the initial volume of nitrogen gas is given by:
V_initial = V_cylinder.
Substituting the values we've calculated earlier:
V_initial = 9.47 * 10^(-4) m^3.
Therefore:
V_final = V_plastic bag + V_cylinder
= V_initial + V_cylinder
= 9.47 * 10^(-4) m^3 + 9.47 * 10^(-4) m^3
= 1.894 * 10^(-3) m^3.
Finally, we need to convert the volume from cubic meters to liters. There are 1000 liters in 1 cubic meter.
1.894 * 10^(-3) m^3 * 1000 L/m^3 = 1.894 L.
Therefore, the final volume of nitrogen gas, including what's collected in the plastic bag and what's left over in the cylinder, is approximately 1.89 liters, rounded to two significant digits.