Question

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.

Answer 1

Answer:

work done in the process is 0 J.

Explanation:

Given :

Initial pressure of system , $p_i=10\ kPa.$

Final pressure of system , $p_f=20\ kPa.$

Volume of system , $V=5\ m^3.$

We know, In thermodynamics work done is defined as :

$W=P\Delta V$

Here , P is pressure and $\Delta V$ is change in volume.

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

Therefore, change in volume , $\Delta V=0.$

Putting value of $\Delta V=0$ in above equation we get ,

W = 0 J.

Therefore , work done in this process is 0 J.

Hence, this is the required solution.