5500 J of heat were added to three moles of a monoatomic gas. During this process the temperature decreased from 540 K to 350 K. What is the work performed by the gas during this process

The work performed ≅ 1.26 × 10 ⁴ J

Explanation:

The internal energy change of a monoatomic ideal gas is represented by the equation:

ΔU = n*Cv*ΔT

where;

the heat capacity Cv for an ideal monoatomic gas is (3/2) R

Then;

ΔU = n × (3/2) R × ΔT

ΔU = 3 × (3/2) × 8.314 ×( 540-350)

ΔU = 7108.47 J

Using the 1st law of thermodynamics;

The work performed = Heat added + the internal energy change

The work performed = 5500 J + 7108.47 J

The work performed = 12608.47

The work performed ≅ 1.26 × 10 ⁴ J

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.