Question

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?

Answer 1

W= - 7470 J

Step-by-step 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.