Question

A 3 m³ rigid tank contains nitrogen gas at 500 kPa and 300 K. Now the heat is transferred to the nitrogen in the tank and the pressure of nitrogen rises to 800 kPa. The work done during this process is:
1. 500 kJ
2. 1500 kJ
3. 0 kJ
4. 900 kJ

Answer 1

Correct Answer - Option 3 : 0 kJ

Concept:

From the first law of thermodynamic, the heat supplied to the system can be utilized to change the internal energy of the system and help the system to do the work. Mathematically it can be expressed as,

δQ = dU + δW

Calculation:

Given:

V = 3 m³, P₁ = 500 kPa, T = 300 K, P₂ = 800 kPa

From First of Law Thermodynamics:

δQ = dU + δW

where δQ = heat interaction with the system and surrounding, dU = change in Internal energy of the system, δW = PdV = work-interaction by the system.

In the question, it is given that rigid container of the nitrogen gas therefore the change in the volume will be zero.

δW = PdV

δW = 0

So work done will be zero.

The heat supply to the system can be utilized to change the internal energy of the gas and because of that only temperature and pressure of the gas will change.