Question

The cyclic integral of (δQ−δW) for a process is

Answer 1

According to first law of thermodynamics :When a closed system undergoes a cycle then the cyclic integral of heat is equal to the cyclic integral of work.  

It can be written as:

$\partial Q = \partial W$  

Or it can be written as:

$Q_{A_{1-2}}+Q_{B_{1-2}}=W_{A_{1-2}}+W_{B_{1-2}}$  

That is:

$\partial Q - \partial W =0$

So,

$(\partial Q - \partial W) =0$

Answer 2

Answer:

positive,negative or zero .

Explanation:

for a cycle it is zero...for a process it can be +,-,0