Question

A system undergoes a two-step process. In step 1, it absorbs 50. J of heat at constant volume. In step 2, it releases 5J of heat at 1 atm. as it returned to its original internal energy. Find the change in the volume of the system during the second step and identify it as an expansion or compression.

Answer 1

Answer:

$\Delta V=0.44\ L.$

Expansion.

Explanation:

In step 1:

Since, volume is constant.

Therefore, Work Done , w=0.

q= 50\ J.

We know equation of thermodynamics,

$\Delta U=q+w=0+50=50\ J$  

Step 2:

Heat is released therefore, $q'=-5\ J.$

As system returned to its initial state.

Therefore, $\Delta U'=-50\ J.$

So, Work Done, $w'=\Delta U'-q'=-50 -(-5)\ J=-45\ J.$

Also, $w'=P\times \Delta V=101325\ atm\times \Delta V.$

Therefore, $\Delta V=0.44\ L.$

Since, $\Delta V$ is positive.

Therefore, it is expansion.

Hence, this is the required solution.