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Murljashka [212]
3 years ago
8

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.
Chemistry
1 answer:
ycow [4]3 years ago
6 0

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.

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