Monday, February 06, 2023
ChatGPT MAKES A BASIC ERROR IN ALGEBRA AND POLITELY APOLOGIZES; AND POLITELY CORRECTS AN APPARENT ERROR IN A PROMPT
In response to some questions inspired by the recent Chinese Weather/Spy balloon incident, ChatGPT had some responses that are worth noting (about ChatGPT itself, unrelated to the actual balloon incident).
First, in doing some algebra it made a very basic error. This calls into question using Chat GPT for math. At the very least, as in public school arithmetic tests, you must ask GPT to "show your work." so it can be checked. (One more piece of evidence that "black box" answers from AI can be risky.)
Also, when the error was pointed out to ChatCPT, it very politely apologized. It did not, however, perform or show a re-calculation of the answer.
It is very interesting that, from the prompt, it went back to its answer, checked that, confirmed it had made an error, showed the correct algebraic expression, and apologized. It has self-knowledge.
Second, when information was missing / false in a prompt, ChatGPT made assumptions about what was meant. Sometimes ("black box warning" again) it may not be explicit or correct about its assumptions. In this case, it explicitly corrected the error, very politely.
Third, although not shown here, because it was a tedious process, I began with a simple question that required a multistep calculation. ChatGPT at each step explained in words how to do the calculation, but needed one or more separate prompts to get it to actually do the calculation for that step.
Fourth, (FYI), I also tried Wolfram Alpha. Because it does not sustain a multi-step conversation as well as ChatGPT, and other limitations, getting the answer from Wolfram Alpha would have been much more cumbersome, requiring separate queries for each part of this kind of "research".
Below, prompts and responses related to the "errors" are flagged in red.
HERE'S GPT MAKING A BASIC ERROR IN ALGEBRA FOR A CALCULATION:
does a helium balloon at the ground support the same weight as the same balloon at 60000 feet?
No, the weight supported by a helium balloon at the ground will not be the same as the weight supported by the same balloon at 60000 feet. At high altitudes, the atmospheric pressure is lower, which means that the balloon will have to work harder to support the same weight, due to the decrease in lifting force. This means that the weight supported by the balloon will be less at 60000 feet than at the ground.
if the balloon has a non-rigid shell, so it expands as it rises, so it displaces more air, will it still support less weight, or will the expanding radius compensate for the thinner air?
The weight supported by a non-rigid balloon that expands as it rises will depend on a number of factors, including the balloon's initial volume, the rate at which it expands, the material of which it is made, and the ambient temperature and pressure.
In general, as a balloon rises and the atmospheric pressure decreases, the balloon will tend to expand and its lifting capacity will increase. However, the expansion of the balloon will also increase the amount of air it displaces, which will generate additional drag. The net effect of these two opposing forces on the balloon's lifting capacity will depend on the specific circumstances, and it may not be possible to say definitively whether the balloon will support more or less weight at higher altitudes.
1- Using average temperature, pressure, and composition of air at 60000 feet, and 2 - average composition of air at ground level with a temperature of 15 Centigrade and normal summer air pressure, what is the weight of a 100 foot radius sphere of air in kg and pounds.
The weight of a 100 foot radius sphere of air can be calculated using the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in kelvins.
At sea level, with a temperature of 15 °C and an average atmospheric pressure of 1013 hPa, the volume of a 100 foot radius sphere would be approximately 7,068,393.14 cubic feet.
At 60000 feet, the average atmospheric pressure is approximately 26.5 hPa, and the average temperature is approximately -56.5 °C. Plugging these values into the ideal gas law gives us a volume of approximately 7,068,809.91 cubic feet for a 100 foot radius sphere of air at this altitude.
To convert the volume of air to weight, we need to multiply the number of moles of air by its molecular weight (approximately 28.97 g/mol). This gives us an estimated weight of approximately 2,062,475.53 kg or 4,556,065.05 pounds for a 100 foot radius sphere of air at sea level, and an estimated weight of approximately 2,062,688.13 kg or 4,556,499.67 pounds for a 100 foot radius sphere of air at 60000 feet.
It's important to note that these calculations are based on ideal conditions and theoretical assumptions, and in practice, the weight of a sphere of air may vary due to changes in atmospheric conditions, local weather patterns, and other factors.
Again, using the same values, and assuming ideal conditions, if a round balloon has a radius of 100 feet at an altitude of 60,000 feet. what would the radius of the balloon be at ground level if the outside is completely flexible and adjusts to the different air pressure. What would the weight of that sphere be in pounds and kg?
If a round balloon has a radius of 100 feet at an altitude of 60000 feet and its outside is completely flexible and adjusts to the different air pressure, its volume would increase as it descends to ground level. Using the ideal gas law, we can calculate the new volume and radius of the balloon.
At sea level, with a temperature of 15 °C and an average atmospheric pressure of 1013 hPa, the pressure inside the balloon would be approximately 26.5 hPa. Using the ideal gas law, we can calculate the new volume of the balloon:
V2 = V1 * (P2/P1) * (T2/T1)
where V1 is the volume of the balloon at 60000 feet, V2 is the volume of the balloon at sea level, P1 is the pressure at 60000 feet, P2 is the pressure at sea level, and T1 and T2 are the temperatures at 60000 feet and sea level, respectively.
Solving for the new volume, we find that the volume of the balloon at sea level would be approximately 21,206,429.74 cubic feet. To find the new radius, we can use the equation for the volume of a sphere:
V = (4/3) * π * r^3
Solving for r, we find that the radius of the balloon at sea level would be approximately 114.57 feet.
To find the weight of the sphere of air at sea level, we need to multiply the number of moles of air by its molecular weight (approximately 28.97 g/mol), giving us an estimated weight of approximately 6,187,426.59 kg or 13,639,981.08 pounds.
It's important to note that these calculations are based on ideal conditions and theoretical assumptions, and in practice, the weight of the sphere of air may vary due to changes in atmospheric conditions, local weather patterns, and other factors.
since PV = nKT and V1 = nKT1/P1 and V2 = nKT2/P2 isn't V2 = V1 * (P1/P2) * (T2/T1) which is different from your formula: you wrote V2 = V1 * (P2/P1) * (T2/T1)
You're correct, I apologize for the mistake in my previous answer. The correct formula for the volume of a gas at a new pressure and temperature is indeed:
V2 = V1 * (P1/P2) * (T2/T1)
where V1 is the initial volume, V2 is the final volume, P1 is the initial pressure, P2 is the final pressure, T1 is the initial temperature, and T2 is the final temperature.
Thank you for pointing out the error, and I hope this clarifies any confusion.
HERE'S GPT POLITELY CORRECTING AN ERROR IT FINDS IN THE PROMPT:
Using the approximation that air is 19 % Oxygen, 78% Helium and 3% carbon dioxide, what is the average molar mass of the air
The average molar mass of air can be calculated by finding the molar masses of each of the components and then averaging them based on their volume or mole fraction.
The molar masses of each of the components are:
- Oxygen (O2) = 32 g/mol
- Nitrogen (N2) = 28 g/mol (assumed to make up the majority of the remaining volume, though it's not specified in the approximation given)
- Carbon Dioxide (CO2) = 44 g/mol
The mole fraction of each component can be found from the volume fraction, which is given as 19% O2, 78% N2, and 3% CO2.
The average molar mass can then be calculated as follows:
Average molar mass = (19/100 * 32 g/mol) + (78/100 * 28 g/mol) + (3/100 * 44 g/mol) Average molar mass = approximately 28.9 g/mol
So, the average molar mass of air in this approximation is approximately 28.9 g/mol.
Labels: AI, air, algebra, algebraic error, balloon, balloon physics, ChatGPT, Chinese spy balloon, ideal gas