A glass bulb with radius 15 cm is filled with water vapor at a temperature of 500 K. On the exterior of the bulb a spot with area 2.0 mm^2 is placed in contact with liquid nitrogen. This causes the temperature of the spot to drop well below 0◦C. When the gaseous water molecules strike this area from inside they immediately freeze effectively removing them from the gas. Determine how much time it takes for the pressure in the bulb to reduce to half its original value. Assume the temperature of the gas in the container remains at 500 K
To determine the time it takes for the pressure to reduce to half its original value, we need to use the ideal gas law.
The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
In this case, we are given that the temperature remains constant at 500 K. Therefore, we can rewrite the ideal gas law as P1V1 = P2V2.
Let's denote the initial pressure as P1 and the final pressure as P2. We want to find the time it takes for P2 to be half of P1.
Since the volume V of the gas remains constant, we can simplify the equation as P1 = P2 * (V2/V1). But we are interested in P2 being half of P1, so we can rewrite the equation as P1/2 = P2 * (V2/V1).
Now, we need to determine the ratio V2/V1. The volume V of a sphere is given by V = (4/3) * π * r^3, where r is the radius.
Given that the radius of the glass bulb is 15 cm, or 0.15 m, the volume V1 of the initial gas is (4/3) * π * (0.15^3) m^3.
When the water vapor freezes, the gaseous water molecules are removed from the gas, effectively reducing the volume of the gas. Let's assume that the gaseous water molecules that freeze occupy the same volume as the spot of liquid nitrogen. Given that the area of the spot is 2.0 mm^2, or 2.0 x 10^-6 m^2, the volume V2 is equal to the area multiplied by the thickness of the layer of gas (considered negligible).
Now, we can calculate the ratio V2/V1 as V2/V1 = (2.0 x 10^-6 m^2) / [(4/3) * π * (0.15^3) m^3].
Next, we substitute the ratio V2/V1 and the desired final pressure P1/2 into the equation P1/2 = P2 * (V2/V1), and solve for P2:
P1/2 = P2 * (V2/V1)
P2 = (P1/2) * (V1/V2)
Finally, we need to determine the time it takes for P2 to reach this value. The rate of change of pressure with respect to time is given by dP/dt = -k * A, where dP/dt is the rate of change of pressure, k is the proportionality constant, and A is the surface area.
In this case, we are given that the pressure reduction is due to the freezing of gaseous water molecules. The water molecules freeze when they strike the spot of liquid nitrogen. Thus, the rate of change of pressure is proportional to the number of water molecules striking the spot in a given time, which is directly proportional to the surface area A of the spot.
We can rewrite the equation as dP/P = -k * dt/A, where dP/P is the fractional change in pressure and dt is the time interval.
In order to integrate the equation, we assume that the temperature remains constant so that the volume and pressure are inversely proportional. Therefore, dP/P = -k * dt/V, where V is the volume.
By integrating the equation, we get ln(P2/P1) = -k * t/V, where ln is the natural logarithm, k is the proportionality constant, t is the time, and V is the volume.
To find the time t, we rearrange the equation as t = -(V/k) * ln(P2/P1).
We now have all the information needed to calculate the time it takes for the pressure to reduce to half its original value. However, we need the value of the proportionality constant k, which can only be obtained experimentally. Once we have the value of k, we can substitute it along with the given values of P1, P2, and V into the equation to determine the time t.