Question

The pressure on the block of copper of 1 kg is increased from 20 bar to 800 bar in a reversible process maintaining the temperature constant at 15°C. Determine the following : 

(i) Work done on the copper during the process, 

(ii) Change in entropy, 

(iii) The heat transfer, 

(iv) Change in internal energy, and 

(v) (c_p – c_v) for this change of state. 

Given : β (Volume expansitivity = 5 × 10^–5/K, K (thermal compressibility) = 8.6 × 10^–12 m²/N and v (specific volume) = 0.114 × 10^–3 m³/kg.

Answer 1

(i) Work done on the copper, W : 

Work done during isothermal compression is given by

W = \(\int_1^2pdv\)

The isothermal compressibility is given by

Since v and K remain essentially constant

The negative sign indicates that the work is done on the copper block. 

(ii) Change in entropy : 

The change in entropy can be found by using the following Maxwell relation :

Integrating the above equation, assuming v and β remaining constant, we get

s₂ – s₁ = – vβ (p₂ – p₁)_T 

= – 0.114 × 10^–3 × 5 × 10^–5 [800 × 10⁵ – 20 × 10⁵] 

= – 0.114 × 10^–3 × 5 (800 – 20) = – 0.446 J/kg K.

(iii) The heat transfer, Q : 

For a reversible isothermal process, the heat transfer is given by : 

Q = T(s₂ – s₁) = (15 + 273)(– 0.4446) = – 128 J/kg

(iv) Change in internal energy, du : 

The change in internal energy is given by :

du = Q – W 

= – 128 – (– 3.135) = – 124.8 J/kg.

(v) c_p – c_v : 

The difference between the specific heat is given by :

c_p – c_v = \(\cfrac{\beta^2Tv}K\)

= \(\cfrac{(5\times10^{-5})^2\times(15+273) \times0.114 \times10^{-3}}{8.6\times10^{-12}}\) 

9.54 J/kg K