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**Question**

A mass of 1.5 kg of air at 120 kPa and 24 C is contained in a piston-cylinder device. The air is compressed to a final pressure of 600 kPa. During the process heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the work input during the process.

**My Solution**

I've solved the problem by the isothermal PV equation the answer is -206 kJ but have some problems with the substituting the result in the first law.

We know that $$ E_(In) - E_(Out) = \Delta U$$ and also $$ Q(In) - Q_(Out) - W_b = \Delta U$$

Ok, there is no heat input in the question so Q(in) = 0 and the Q(out) is the energy that is transferred from the air. W (b) is the boundry work that we found -206 kJ which means work input and finally because the process is isothermall Delta U = 0.

Now lets substitute, $$ - Q_(Out) = -206 kJ $$ If the signs are correct what exactly that means.

We have done 206 kJ work on the system and to be able to keep the temperature constant, there must be taken 206 kJ energy from the air? I am confused here can you please explain the result.

Thanks,

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