The beaker in FIGURE P19.45, with a thin metal bottom, is filled with 20 g of water at 20°C. It is brought into good thermal contact with a 4000 cm^3 container holding 0.40 mol of a monatomic gas at 10 atm pressure. Both containers are well insulated from their surroundings. What is the gas pressure after a long time has elapsed? You can assume that the containers themselves are nearly massless and do not affect the outcome.

Question

Accepted Answer

Hey, everyone. So this problem is dealing with heat transfer and the ideal gas law, let's see what it's asking us. So we have a scientist conducting a thermal equilibrium experiment and we have this double walled canister that has a thin metallic partition that keeps the water and the gas separate. So on one side, we have 45 g of water at 25 degrees Celsius. And then on the other side, we've got 5000 cubic centimeters of half mole of a mono atomic gas, which is initially at 12 atmospheres of pressure. So that partition allows for effective thermal contact between um the two, the liquid and the gas. And then the whole container ensures proper insulation, which tells us that we have conservation of energy. We have a closed system that we're dealing with here. So after a substantial amount of time has passed, we are asked to find the new gas pressure. So after it has reached thermal equilibrium, what is that gas pressure? And um our last assumption that we're told is that the container has negligible mass and it does not influence the final result. And so our multiple choice answers here are a 3.39 times 10 to the fifth pascals B 2.41 times 10 to the sixth pascals C 2. times 10 to the fifth pascals or D 3.82 times 10 to the six pascals. And so the first thing we can do here is recall that our ideal gas law is given by PV equals N RT. And so this monotonic gas is an ideal gas initially and in our final condition. And so our volume doesn't change between initial and final, the N number of moles doesn't change. And then our R is our gas constant, it's just a constant. And so that tells us that the relationship between our initial condition and our final condition is going to be our initial pressure divided by our initial temperature is equal to our final pressure divided by our final temperature. And we are looking for this final pressure. So when we rewrite the equation, we get P F is equal to P I multiplied by T F divided by T I. Now, when we look at what was given to us in the problem, we do have our initial pressure, but we need to find our initial temperature and our final temperature in order to solve our hour final question. So we can go back to this ideal gas log looking at our initial conditions to find our initial temperature. So that looks like P I V or initial pressure, multiply buyer volume is equal to N RT I T initial. And so we have all of these terms. So we can rearrange the equation isolating our initial temperature. And that's P I multiplied by V divided by N multiplied by R is equal to our initial temperature. And so plugging in, we have 12 atmospheres, we need to get that in standard units. So we can recall the conversion factor between atmospheres and pascals is 1.13 times 10 to the fifth pascals per atmosphere. And then multiplied by our volume. 5000 cubic centimeters is the same as five times 10 to the negative three cubic meters. Again, just keeping everything in standard units. And then, and our number of moles is going to be just 0. moles and then our gas constant we can recall is 8. joules per mole. How then? And so that initial gas temperature when we plug that in, we get 1462 degrees Kelvin. Ok. So now we need to find our final temperature which is going to be the final temperature of both the water and the gas. And so knowing that we have this insulated closed system is a hint, big hint for us to use our conservation of energy equation which tells us that the heat energy of the gas plus the heat energy of the water is equal to zero where our heat energy is given by the equation mass multiplied by specific heat capacity multiplied by our change in temperature. Now, another way that we can use um this equation instead of mass and specific heat capacity, an equivalent would be our molar mass and our molar heat capacity and then still multiplied by that um delta T. And so that is a, another way we don't have the mass of the gas. We have the molar mass or the number of moles of the gas. We can use that equation. Excuse me. So this looks like our mold of gas multiplied by our molar heat capacity of gas multiplied by our final temperature. And the final temperature is going to be again, the same between the gas and the water minus our initial gas temperature is equal to negative mass of the water multiplied by the specific heat capacity of the water multiplied by RT F minus initial temperature of the water. Let me see if that a little bit clear. And so isolating for our final temperature, which is what the problem is asking us for. That looks like the mass of the water multiplied by the specific heat capacity of the water multiplied by our initial temperature of the water plus our moles of gas multiplied by the molar heat capacity of gas multiplied by our initial gas temperature, all divided by our moles of gas multiplied by our specific heat capacity or, or Mueller, excuse me, our Mueller heat capacity of gas plus our mass of water multiplied by our specific the capacity of water. And then from there, we actually have all of the terms that we need to solve for that final temperature. And then that's our last step before we can plug it in to solve for our final pressure. So our mass of the water is given and the problem as um grams, yeah, 45 g. And so we can rewrite that as 450.45 kg multiplied by our specific heat capacity of water. We can recall that's 4190 Jews per kilogram, Kelvin. And then multiplied by our initial water temperature was given as 25 degrees Celsius. And so we can just add 273 degrees to get that conversion between Celsius and Kelvin. So it would be 25 plus 273 Kelvin. And then plus our gas terms, we've got 0.5 moles multiply by our molar heat capacity. So we can recall that our molar heat capacity at a constant volume of a monatomic gas is 12. jewels per mole. And then our initial gas temperature, we just solve horror 1462 K. And then all of that divided by moles and gas 0.5 moles multiplied by 12.5 jewels per mole plus our massive water. Again, 0.45 kg multiplied by our specific key capacity of water, 4190 Jews per kilogram kit. And we flood that in and we get a final temperature of 335 Kelvin. And so now we have everything we need to go back up to that first equation to solve for our final pressure. So our final pressure is equal to our initial pressure. Again, that was 12 atmospheres getting that in standard units multiplied by 1.13 times 10 to the fifth past gas per atmosphere multiplied by our final temperature. She just saw for 335 Kelvin divided by our initial temperature 1462. Kelvin, we plug that in and we get a final pressure of 2.79 times 10 to the fifth hot scales. And so that is the final answer to this problem. And when we look at our multiple choice answers, it aligns with answer choice C so that's all we have for this one. See you in the next video.

Verified Solution

Key Concepts

Thermal Equilibrium

Ideal Gas Law

Monatomic Gas Properties