Question

0.44 kg of air at 180°C expands adiabatically to three times its original volume and during the process, there is a fall in temperature to 15°C. The work done during the process is 52.5 kJ. Calculate c_p and c_v.

Answer 1

Mass of air, m = 0.44 kg 

Initial temperature, T₁ = 180 + 273 = 453 K

Ratio = V₂/V₁ = 3 

Final temperature, T₂ = 15 + 273 = 288 K 

Work done during the process, W_1–2 = 52.5 kJ 

c_p = ?, c_v = ? 

For adiabatic process, we have

or Taking log on both sides, we get 

log_e (0.6357) = (γ – 1) log_e (0.333) 

– 0.453 = (γ – 1) × (– 1.0996)

∴ γ = \(\cfrac{0.453}{1.0996}\) + 1 = 1.41

Also, c_p/c_v = γ = 1.41

Work done during adiabatic process,