Question

In an isothermal process 150 J of work is done on an ideal gas, calculate $Q$ and $\Delta U$.

Answer 1

As the process is isothermal, thus the change in internal energy will be zero, i.e.,

$\Delta U=0$

Now from first law of thermodynamics,

$\Delta U=q+w$

$\because \Delta U=0$

$\therefore q=-w$

Given that, the work is done on the gas.

$\therefore w=+150J$

Therefore,

$q=-w=-150J$