Question

A gas within a piston–cylinder assembly undergoes a

Answer 1

Answer:

$W_{net}= -280.5 kJ$

$Q_{23}= 80.0 kJ$

Explanation:

The net work of the cycle is the sum of works in each process.

For the first process 1-2: Let's apply the work definition.

$W=\int pdV$ (1)

Now, we need the pressure. We know that pV=C, where C is a constant. Then  $p_{1}V_{1}=10^{5}2=2*10^{5} [J]$

So $p=\frac{2*10^{5}}{V}$

Let's put p in (1):

$W_{12}=\int \frac{2*10^{5}}{V}dV=2*10^{5}\int \frac{1}{V}dV$

$W_{12}=2*10^{5}(ln(V_{2})-ln(V_{1}))=2*10^{5}(ln(0.2)-ln(2))=-460.5 kJ$

Using the first law of thermodynamics we can find Q.

$Q_{12}-W=\Delta U$

$Q_{12}=W+\Delta U=-460.5 + 100= -360.5 kJ$

Second process 2-3

In this case, we have a constant volume, so the work done here is 0.

$W_{23}=0$

Third process 3-1

$W_{31}=\int pdV=p\int dV=p(V_{2}-V_{1})$

$W_{31}=10^{5}(2-0.2)=180 kJ$

Finally, the net work is:

$W_{net}=W_{12}+W_{23}+W_{31}= -280.5 kJ$

By the conservation of energy:

$(Q_{12}+Q_{23}+Q_{31})-(W_{net})=\Delta E=0$

Because there is no change in total energy.

So:

$Q_{23}=W_{net}-Q_{12}= 80.0 kJ$

It is a refrigerator because the net work is negative, it means it consumes energy.

I hope it helps you!