Question

Derive the relationship between work for a reversible reaction and the charge in volume during comoression and expansion.

Answer 1

1. During expansion, work is done by the system; since V₁ > V₁, the sign obtained for work will be negative. 

2. During compression, work is done on the system; since V < V., the sign obtained for work will be positive. 

3. If the pressure is not constant, but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, dV. 

4. We can calculate the work done on the gas by the relation,

\(w =\int\limits_{V_i}^{V_f}PdV ...(1)\)

5. In a compression process, P_ext the external pressure is always greater than the pressure of the system. 

i.e., P_ext = (P_int + dp). 

In an expansion process, the external pressure is always less than the pressure of the system 

i.e., P = (P – dp).

6. In general case, we can write, P_ext = (P_int ± dp). Such processes are called reversible processes. For a compression process, work can be related to internal pressure of the system under reversible conditions by writing equation.

Since dp.dv is very small, we can write,

7. If V_f > V_i (expansion), the sign to work done by the process is negative. 

If V_f < V_i (compression), the sign to work done by the process is positive.