I once was given a metal whiskey flask for being in my cousin’s wedding. When it got dented, I put in 1 gram of H2O (initially 21°C, just like the air filling the rest of the flask) and heated it up to boil the water. It popped the dent out.
Assume the flask had a rectangular cross section, with a height of 15 cm, a width of 10 cm, and a thickness of 2 cm. Assume that in order to pop the dent out, an outward pressure difference of 1.0E6 Pa was required (that is, the outward pressure on the flask minus the inward pressure on the flask was 1.0E6 Pa). How much heat had to be added to the contents of the flask?
At the instant the dent popped out, was the gas inside the flask doing positive work, or was positive work being done to the gas?
the metal was moving in the direction the gas was pushing. That is positive work.
As for the first part, figure out how many mols of water vapor you have and how many mols of air, Then do PV = nRT to get partial pressures of each. That final pressure inside is ten times atmospheric plus 1 atm.
To calculate the amount of heat that had to be added to the contents of the flask, we need to consider the process of heating the water and the subsequent expansion of the gas.
First, let's calculate the volume change of the gas when the dent pops out. Given the rectangular cross-section dimensions of the flask (height = 15 cm, width = 10 cm, thickness = 2 cm), the initial volume of the flask can be calculated as follows:
Initial volume = height x width x thickness = 15 cm x 10 cm x 2 cm = 300 cm³
Since the flask was initially filled with air at the same temperature as the water (21°C), we can assume the air inside the flask also reached the boiling point of the water (100°C) after the water was heated.
Now, let's calculate the final volume of the flask when the dent pops out. We know that the outward pressure difference required to pop the dent is 1.0E6 Pa. Since the flask initially had atmospheric pressure inside, the final pressure inside the flask would be atmospheric pressure (around 1.0E5 Pa).
From the ideal gas law (PV = nRT), we can assume the number of moles (n), the universal gas constant (R), and the temperature (T) of the gas remains constant during this process. Therefore, we can write:
Initial pressure x initial volume = final pressure x final volume
1.0E5 Pa x 300 cm³ = 1.0E6 Pa x final volume
final volume = (1.0E5 Pa x 300 cm³) / 1.0E6 Pa = 30 cm³
The change in volume is given by:
Volume change = final volume - initial volume = 30 cm³ - 300 cm³ = -270 cm³
Since the dent popped out, the volume increased, so the negative sign indicates the change in volume.
Now, to calculate the amount of heat added to the contents of the flask, we can use the equation:
Heat = m x c x ΔT
Where:
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
Given that 1 gram of water was added, the mass of the water is 1 gram.
The specific heat capacity of water is approximately 4.18 J/g°C.
The change in temperature is the difference between the initial and final temperatures, which is the boiling point of water (100°C) minus the initial temperature (21°C).
ΔT = 100°C - 21°C = 79°C
Plugging in these values, we can calculate the heat added:
Heat = 1 gram x 4.18 J/g°C x 79°C = 329.42 J
Therefore, the amount of heat that had to be added to the contents of the flask is approximately 329.42 J.
At the instant the dent popped out, positive work was being done to the gas inside the flask. The outward pressure difference caused the gas to expand and push against the dent, resulting in the dent popping out. Work was done on the gas by the external pressure to displace it and restore the flask's original shape.