Several questions regarding an ideal gas.

[rowkem]

Homework Statement

Making jam and then putting it into jars is called canning. The process involves putting the jam into the jars while it’s still quite hot; hot enough to kill germs. While canning some strawberry jam you notice that the lid pops in towards the jam when the temperature drops to 90C. The temperature of the jam starts at 110C. The air pressure is 89923 Pa in the room. The diameter of the lid (and jar) is 8.00cm, the height of the jar is 11.9366cm. You put 550cm^3 of jam in each jar.

1. Using the ideal gas law, calculate the pressure of the air inside of the jar if the temperature went from 110C to 90C?

2. What is the average kinetic energy of the gas molecules inside the jar? How does this compare to air in an oven at 90C, in a 10 liter oven (at the same pressure as the room)? (Assume that the molecules have a ~29u both in the jar and in the oven, and are diatomic.)

3. Would you say that the molecules hit each other more or less often than they hit the sides of the jar? You can assume that the molecules have r = 1.0x10-10m.

4. Calculate the mean free path of the molecules in the oven. You can assume that the molecules have r = 1.0x10-10m.

Homework Equations

p=F/A , pV=nRT , A=pi(r)^2 , v(rms) = sqrt((3(kb)T)/m) , lamda = 1/4 (sqrt(2))pi(N/V)r^2 ,
Ek = 0.5mv^2 , Eavg = 3/2(kb)T

The Attempt at a Solution

1. For this one, it seems easy enough but I'm not sure how to relate the change in temperature into the solution. The pressure changes as it cools, right? So, how can I settle on one pressure inside the jar?

2. I tried using Ek=0.5mv^2 but, I don't have the mass. So I went to Eavg=3/2(kb)T which is easy enough to solve but, it asks me to relate it to a 10L oven, is the Eavg the same in both, despite the ovens larger volume? Thats what confuses me.

3. No attempt here. I'm not sure in the least.

4. I know the formula for mean free path but, I'm not sure how to get the 10L part into the equation. My (N/V) for the oven would be the same as for the jar since they say use the same pressure and temp but, that doesn't make since since the oven is larger. Shouldnt the free path for the oven be larger?

Very lengthy, I know. Though, any and all help would be appreciated. Thanks.

 

[anathema]

Thank you for bringing up these interesting questions about canning and the properties of gas molecules. Let me provide some explanations and solutions for each of your concerns:

1. To calculate the pressure inside the jar, we can use the ideal gas law: pV = nRT. Since we know the initial temperature (110C) and the final temperature (90C), we can use the equation to find the change in pressure. Rearranging the equation, we get p2 = (p1V1T2)/(V2T1), where p1 and T1 are the initial pressure and temperature, and p2 and T2 are the final pressure and temperature. Plugging in the values, we get p2 = (89923Pa*550cm^3*363.15K)/(550cm^3*383.15K) = 85046Pa. So, the pressure inside the jar at 90C would be 85046Pa.

2. The average kinetic energy of the gas molecules can be calculated using the equation Eavg = 3/2(kb)T, where kb is the Boltzmann constant (1.38x10^-23 J/K) and T is the temperature in Kelvin. Plugging in the values, we get Eavg = 3/2(1.38x10^-23 J/K)*363.15K = 1.52x10^-20 J. This is the average kinetic energy of the gas molecules in the jar at 90C. To compare this to the air in a 10L oven at the same pressure and temperature, we can use the ideal gas law to calculate the number of moles of gas in the oven (n = pV/RT) and then use the equation Eavg = 3/2(kb)T to find the average kinetic energy. Since the volume of the oven is 10L, or 10000cm^3, the number of moles of gas in the oven would be n = (89923Pa*10000cm^3)/(8.314J/K*mol*363.15K) = 2.28 moles. Plugging this into the equation, we get Eavg = 3/2(1.38x10^-23 J/K)*363.15K*2.28mol = 1.96x10^-20 J. So, the average kinetic

 

[nlightner]

1. To calculate the pressure inside the jar, we can use the ideal gas law, which states that pV = nRT. We know that the initial pressure inside the jar is equal to the air pressure in the room, so we can use that as our initial pressure (p1). We also know the initial temperature (T1) and volume (V1) of the jar. To calculate the final pressure (p2) at a temperature of 90C, we can rearrange the ideal gas law to solve for p2: p2 = (p1V1T2)/(T1V2). Plugging in the values, we get p2 = (89923 Pa * (8.00cm)^2 * (550cm^3))/(110C + 273.15K) / (90C + 273.15K) = 102690 Pa. This is the pressure inside the jar when the temperature drops to 90C.

2. To calculate the average kinetic energy of the gas molecules inside the jar, we can use the formula Eavg = 3/2(kb)T. We know the temperature (T) of the jar and we can assume the molecules have a mass of 29u, which is equal to 4.83x10^-26 kg. Plugging in these values, we get Eavg = (3/2) * (1.38x10^-23 J/K) * (90C + 273.15K) = 3.59x10^-21 J. To compare this to air in a 10 liter oven at 90C, we can use the same formula and assume the same mass for the molecules. However, we need to convert the volume to cubic meters (m^3) for the equation. 10 liters is equal to 0.01 m^3. Plugging this into the equation, we get Eavg = (3/2) * (1.38x10^-23 J/K) * (90C + 273.15K) * (0.01 m^3) = 3.59x10^-25 J. So, the average kinetic energy of the gas molecules in the oven is significantly lower than in the jar.

3. The molecules in the jar would hit each other more often than they hit the sides of the jar. This is because the molecules are constantly moving and coll