In an isochoric process, the pressure in a system changes from 10 kPa to 20 kPa at V= 5m3. Calculate the work done in this process according to the sign convention in the class.

Respuesta :

Answer:

work done in the process is 0 J.

Explanation:

Given :

Initial pressure of system , [tex]p_i=10\ kPa.[/tex]

Final pressure of system , [tex]p_f=20\ kPa.[/tex]

Volume of system , [tex]V=5\ m^3.[/tex]

We know, In thermodynamics work done is defined as :

[tex]W=P\Delta V[/tex]

Here , P is pressure and [tex]\Delta V[/tex] is change in volume.

Now, coming back to question , it state that the process is isochoric .

Therefore, change in volume , [tex]\Delta V=0.[/tex]

Putting value of [tex]\Delta V=0[/tex] in above equation we get ,

W = 0 J.

Therefore , work done in this process is 0 J.

Hence, this is the required solution.