# Work in Thermodynamics

Work in thermodynamics is calculated $$W = p \Delta V$$ where p is pressure and V is volume.

I have system where $$p_1 = 105$$ kPa and $$p_2 = 525$$ kPa. Temperature $$T_1=T_2=318.15$$ K and volume $$V_1=1.2 m^3$$ $$V_2=0.25 m^3$$

I want to calculate the work what happens in the process.Do I need to consider the change of the pressure or do I consider only the volume change?

$$W=(p_2-p_1)(V_2-V_1)$$ or $$W=p_2(V_2-V_1$$)

I want to calculate the work what happens in the process.Do I need to consider the change of the pressure or do I consider only the volume change?

While it appears that your system is probably closed, the general answer is it depends on whether your system is a closed system or an open system. That's because there are two general types of possible work (excluding shaft work), depending on whether is system is closed (no mass transfer permitted) or open (permitting mass transfer).

For a closed system, work is referred to as "boundary work", or $$PdV$$ work, which occurs when a force acts through a displacement. For boundary work the general expression is

$$W=\int p\,dV$$

For an open system, we have flow work, or $$VdP$$ work. This is the work needed to push a working fluid into or out of the boundaries of the system against a pressure difference. The general expression for flow work is

$$W=-\int v\,dP$$

The above said, you cannot determine the work done unless you know the process or path that takes you from state 1 to state 2. Or, alternatively, if for a closed system you have enough information to determine the heat transfer $$Q$$ and the change in internal energy $$\Delta U$$ so you can determine the work done $$W$$ by applying the first law, $$\Delta U=Q-W$$. All you have been given are the equilibrium conditions for the two states, not the path between. If the system were an ideal gas, $$\Delta U=0$$ because there is no change in temperature. But you would still need to know $$Q$$. There are an infinite number of possible paths between the two states, each potentially giving a different value of work.

Hope this helps.

The more general form of the expression for work here is given by $$\text dW=p\,\text dV$$

The only time $$W=p\Delta V$$ is valid is if the pressure is constant during the process you want to compute the work for. Therefore, you will need to know how the process moves between the two points in $$p$$ -$$V$$ space. From there you will have to do an integral

$$W=\int p(V)\,\text dV$$

Since this is an isothermal process, if we assume the system is an ideal gas you know what $$p(V)$$ should be. Use this and the above integral to calculate the work done. I will leave that step to you :)

If you are in a non-calculus based class, then your instructor should provide you with an equation for determining work done on/by an ideal gas in an isothermal process. This is something that should be covered in most text books, and I am sure there are numerous examples online.