The temperature of a monatomic ideal gas remains constant during a process in which 7470 J of heat flows out of the gas. How much work (including the proper + or - sign) is done?

W= - 7470 J      

Explanation:

Given that

heat flow out of the system ,Q = - 7470 J

We know that internal energy of the ideal gas depends only on temperature and is given as

ΔU = m Cv ΔT

m=mass

Cv=Specific heat at constant volume

ΔT=Change in the temperature

ΔU=Change in the internal energy

Here given that temperature of the gas is constant ,that is why

ΔT = 0 ⇒ ΔU= 0

We know that ,from first law of thermodynamics

Q= Δ U + W

Q=Heat transfer

W=Work transfer

ΔU=Change in the internal energy

Now by putting the values in the above equation

- 7470 = 0 + W

W= - 7470 J

This applies that work is done on the system.

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

conduction, convection, radiation