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An ideal gas is taken along the path AB (see fig.). If 70 calories of heat is extracted from the gas during the process, calculate the change in internal energy during the process.

My attempt :

Work done in the process is the area of the shaded region which turns out to be 52.5 Joules.

Again, heat extracted,

$$Q=-70 \, \text{cal}= -294 \, J$$

So, from first law of thermodynamics,

$$\begin{align}\Delta U &= Q-W\\ &=(-294-52 5)\, J\\ &= \boxed{-346.5 \, J}\end{align}$$

So, that was my answer. But my teacher says that the answer is -241.5 Joules. For that, he says that the work done should be taken as negative. What can the possible explanation for this ? Where am I wrong ?

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