A meteorological balloon contains 250.0 l he at 22°c and 740.0mmhg if the volume of the balloon can vary according to external conditions what volume would it occupy at an altitude at which the temperature is -52°c and the pressure is 0.750 atm

243.15 l

Explanation:

Using the general gas law

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁T₂/P₂T₁ where P₁, V₁ and T₁ are the initial pressure, volume and temperature of the meteorological balloon and  P₂, V₂ and T₂ are the initial pressure, volume and temperature of the meteorological balloon.

P₁ = 740 mmHg V₁ = 250 l and T₁ = 22°C + 273 = 295 K,

P₂ =  0.750 atm = 0.750 × 760 mmHg = 570 mmHg (since 1 atm = 760 mmHg) and T₂ = -52°C + 273 = 221 K  

V₂ = P₁V₁T₂/P₂T₁ = 740 mmHg × 250 l × 221 K/570 mmHg × 295 K = 243.15 l

V₂ = 243.15 l

So the volume it would occupy at an altitude of temperature -52 °C where it pressure is 0.750 atm  is 243.15 l

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.

the objects have the same charge.