If a balloon with a volume of 10L at 500kpa is heated until the balloon has a temperature of 300°C, 750atm and now has a volume of 100mL. What was the original temperature of the balloon (in °C)

103.6°C

Explanation:

We solve this problem, with the Ideal Gases Law. We know that moles of gas are not changed through time.

P . V = n . R . T

So n (number of moles) and R (Ideal Gases constant) are the same, they are cancelled. In the two different states, we can propose:

P₁ . V₁ / T₁ = P₂ . V₂ /T₂

We need to make some conversions:

100 mL . 1 L/ 1000mL = 0.1L

500 kpa . 1atm / 101.3 kPa = 4.93 atm

300°C + 273 = 573 K

We replace data:

4.93 atm . 10L / T₂ = 750 atm . 0.1L / 573K

4.93 atm . 10L = (750 atm . 0.1L / 573K) . T₂

(4.93 atm . 10L) / (750 atm . 0.1L / 573K) → T₂

T₂ = 376.65 K

We convert K to °C →  376.65 K - 273 = 103.6 °C

jjj89t'ji; it hink the answer is your mom is not si

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.