Question

An insulated cylinder of volume capacity 4 m³ contains 20 kg of nitrogen. Paddle work is done on the gas by stirring it till the pressure in the vessel gets increased from 4 bar to 8 bar. Determine : 

(i) Change in internal energy, 

(ii) Work done, 

(iii) Heat transferred, and 

(iv) Change in entropy. 

Take for nitrogen : c_p = 1.04 kJ/kg K, and c_v = 0.7432 kJ/kg K.

Answer 1

Pressure, p₁ = 4 bar = 4 × 10⁵ N/m² 

Pressure, p₂ = 8 bar = 8 × 10⁵ N/m² 

Volume, V₁ = V₂ = 4 m³

and it is constant for both end states.

R = c_p – c_v = 1.04 – 0.7432 = 0.2968 kJ/kg K.

The mass of the gas in the cylinder is given by

(i) Change in internal energy

∆U = (U₂ – U₁) 

= m_cv (T₂ – T₁) = c_v (mT₂ – mT₁) 

= 0.7432 (10781.6 – 5390.8) = 4006.4 kJ.

(ii) Work done, W : 

Energy in the form of paddle work crosses into the system, but there is no change in system boundary or pdv work is absent. No heat is transferred to the system. We have

= 20 × 0.7432 loge 2 = 10.3 kJ/K