Work done in thermodynamics The work done in thermodynamic process is given by the integral of Pdv and also we can write so assuming a quasi static process between two points. But this work is then non dissipative work or work done by conservative force field , then how can it be path dependent ?
 A: Consider the case of an ideal gas satisfying the relation 
$$PV=nRT$$
where $P$ is the pressure, $V$ is the volume, $T$ the temperature and $n$ the number of moles of gas particles. The work done is (assuming an isothermal process) 
$$W=\int_{V_1}^{V_2} PdV=nrT\int_{V_1}^{V_2} \frac{dV}{V}=nRT\ln\left(\frac{V_2}{V_1}\right)$$
At first sight one may see that the work done depends only on the initial and final states of volume and hence is path independent and is a state function. But when you consider a reversible cyclic process and draw a $P-V$ diagram of it, as shown below:  

Suppose at first you go along path 1 from $(P_2,V_2)$ to $(P_1,V_1)$. The area under this graph gives the work done. Now you return along some another path, say, path 2. Here the area under graph gives the work done for the process. As you can see the wok done along the two paths is clearly not the same. The area enclosed by the curve gives the net work done over a complete cycle. i.e., when you return to the original state, the work done is not zero. This means it is not a function of state. This can be proved by first law of thermodynamics which states that:
$$dU=\bar{d} Q-PdV$$  
On completing a cycle, the internal energy returns to the original state so that $dU=0$. This means  
$$\bar{d} Q=PdV$$  
i.e., the work done is equal to the heat added to the system and is hence non-zero (since for an isothermal process heat transfer takes place so as to maintain a constant temperature) which means it is not a function of state.  
Now come back to our ideal gas equation. The above equation states that for a given temperature, work depends on initial and final volumes only along a single path. If you change the path, you change the temperature of the system (since for an isothermal process $PV=constant$). This means the state of the system has changed. Thus there is no possible way to be work independent of the path for a given state of the system.  
For an enlightenment discussion see: 

Introduction to Statistical Mechanics by Roger Bowley,  Mariana Sánchez

A: The work is not done by a conservative field. If the field were conservative the force made by the piston should be the same at a given volume, that is for a given position of the piston, regardless of the state of the gas. But this is not true. For instance, you can add heat at constant volume to the gas, with results in an increase in pressure.  The force that moves the cylinder is now different, and thus the force is not conservative.
A: Work is defined by formula below for any kind of processes not only for quasi static process.
$$\delta w=P\mathrm dv$$
But, what is the $P$?
$P$ is the pressure that resists against the system boundary movement not pressure of the system itself. So, you don't need to have a quasi static process for calculating the work because you don't need pressure of the system.
Work is path dependent because for the same initial and final states, it is possible that its amount be different as you can see in the figure of Unnikrishnan's answer.
