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

po-215 i got it right on the quiz