If irreversible process is quasistatic then can we find it's work using $pdv$ formula? Now it's given on Wikipedia that quasistatic process can be irreversible,so can I find the work done by the closed system undergoing a process using $pdv$ work?...Next, is that work less than the reversible process undergoing the same process? If yes then how and if no then how?
 A: In the wiki article, they discuss friction as a source of irreversibility.  This would typically involve friction between the piston and the cylinder.  If your "system" is defined as the combination of gas plus piston (so that the friction is occurring within the system), then the process experienced by this system is not reversible.  And, of course, the work done by this system on its surroundings quasi statically (at the outside face of the piston) is not PdV.  
But if your "system" is defined as just the gas, then the process experienced by this system is considered reversible in terms of what the gas experiences. (It experiences the exact same deformation and heat history as in a fully reversible process). And, in this case, the work done by the system on its surroundings quasi statically (at the inside face of the piston) is PdV.  And, in this case, the gas is also receiving heat from its surroundings (the piston), assuming all the frictional heat generated goes into the piston, which then transfers heat to the gas (quasi statically).
So, in short, it all depends on what you call your system.  In the first case, all the entropy generation takes place inside the piston.  In the second case, all the entropy generation takes place outside the system.
