The gas left in a used aerosol can is at a pressure of atm at 27 deg C. If this can is thrown into a fire, what is the internal pressure of the gas when it's temperature reaches 927 deg C

P = 4 atm

Explanation:

In this case, we need the initial pressure which you are not providing that. I will use a pressure value of 1 atm to do this, and then, use your own value and replace it in this procedure for an accurate result.

Now, we have an aerosol can with some gas left in there at 1 atm and 27 °C. Then, the can is thrown into the fire, and the temperature raises to 927 °C. The new pressure should be higher.

In this case, the volume of the gas is not being altered, only the temperature and pressure within the can, so, we have a constant volume and we can use the Boyle's law for a constant volume, which is the following:

P₁/T₁ = P₂/T₂

From here, we can solve for P₂:

P₂ = P₁T₂/T₁

Temperature must be at Kelvin so:

T₁ = 27 + 273 = 300 K

T₂ = 927 + 273 = 1200 K

Now, replacing the data we have:

P₂ = 1 * 1200 / 300

P₂ = 4 atm

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j