Question

How can I decide in a thermodinamic process from pressure and/or volume change if work done by the system or work done on the system? Since I know if work done on the system dU=dW+dQ, work done by the system dU=dQ-dW. However for me very difficult to decide whether in the final answer work will be neative or positive. Is it true that if it is compressionw:work done by the system (work negative) and expansion:work done on the systen (work positive)? Does this methology work for heat transfer (Q) and internal energy (U)? When are they positive and when negative?

If it is a cyclic process shall I just simply add these numbers to calculate the final work?

Thank you in advance

Answer 1

In the most general form, the change in internal energy(U) can be written as

dU = TdS+ PdV + dN + vdp + dL + EdP + BdM

Where TdS is heat supplied to the system(Q), -PdV is the mechanical work done on the system other terms are chemical work done(dN) , work done by changing kinetic energy has two parts linear(vdp) and rotational (dL ) electric and magnetic work done are EdP and BdM respectively.

But when only mechanical work is there along with heat transfer then the equation reduces to

dU = TdS+PdV = dW+dQ

or you can write

dW = dU - dQ ## The work done is by two things one by decreasing internal energy and other by heat.

This is the standard first law of thermodynamics. This the only one which should be used by you whatever the case either work done by system or work done on system.

When the sign on right hand side comes positive when calculated from the standard(##) equation then work is done on the system and when it is negative then work is done by the system.

Whenever there is only mechanical work done is there (only the case for which the standard first law is written as discussed above) Then compression work is positive and it is work done on the system whereas expansion work is negative and it is work done by the system.

Yes, this methodology work for heat transfer (Q) and the internal energy(U). Whene heat is given to the system it is positive and when given out by system it is negative by convention. When internal energy increases dU is positive and when it is decreases dU is negative.

You just need to rearrange the terms if you want heat energy you do like  

dQ  = dU - dW

and if you want to change in internal energy then do

dU=dW+dQ

Just note the signs on the right-hand side the equations and don't play with left-hand sides signs