A system executes a cyclic process during which there are two processes as given below:

₁Q₂ = 460 kJ, ₂Q₁ = 100 kJ and ₁W₂ = 210 kJ

What will be work interaction in process ₂W₁?

Concept: 

The first law of thermodynamics is a restatement of the law of conservation of energy. It states that energy cannot be created or destroyed in an isolated system; energy can only be transferred or changed from one form to another.

When heat energy is supplied to a gas, two things may occur:

• The internal energy of the gas may change
• The gas may do some external work by expanding

According to the first law of Thermodynamics:

δQ = δW + ΔU

When a process is executed by a system, the change in stored energy of the system is numerically equal to the net heat interaction minus the net work interaction during the process:

ΔU = δQ – δW where U is the internal energy that is introduced by this law.

For the cyclic process: ΔU = 0

\(\oint \delta Q = \oint \delta W\)

Calculation:

Given that for a cyclic process heat transfers are 1Q2 = 460 kJ, 2Q1 = 100 kJ and 1W2 = 210 kJ

For the cyclic process,

ΔU = 0

\(\oint \delta Q = \oint \delta W\)

∴ 1Q2 +  2Q1 = 1W2 + 2W1

460 + 100 = 210 + 2W1

∴ 2W1 = 350 kJ