Question

A gas is taken along the path AB as shown in figure. If 70 cal of heat is extracted from the gas in the process, calculate the change in the internal energy of the system.
Figure

Answer 1

The change in the internal energy of the system −241.5 J .

Explanation:

It extracts 70 cal of heat from the system.

$\Delta Q=-70 \mathrm{cal}=-(70 \times 4.2) J=-294 \mathrm{J}$  

Step 1:

From the thermodynamics first law  

 $\Delta W=P \Delta V$

If P is considered as the average pressure between points A and B and ΔV the variation of the volume of system from point A to point B

$\begin{aligned}&\Delta W=-\frac{1}{2} \times(200+500) \times 10^{2} \times\left(150 \times 10^{-6}\right)\\&\Delta W=-\frac{1}{2} \times 700 \times 150 \times 10^{-3}\\&\Delta W=-525 \times 10^{-1}=-52.5 J\end{aligned}$

Step 2:

Negative signs are taken here because the final volume is smaller than the initial volume

∆U = ?

∆Q = ∆U + ∆W

∆Q = −294 J

Negative sign here suggests heat is being extracted from the device.

− 294 = ∆U −52.5

∆U = − 294 + 52.5 = −241.5 J

Answer 2

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