Question

A monatomic ideal gas expands slowly to twice its original volume, doing 370 j of work in the process. part a find the heat added to the gas if the process is isothermal.

Answer 1

The first law of thermodynamics says that the variation of internal energy $\Delta U$ of a gas is equal to the amount of heat Q supplied to the gas minus the work W done by the gas:
$\Delta U = Q-W$

The variation of internal energy of a gas is:
$\Delta U = \frac{3}{2} n R \Delta T$
As it can be seen, it depends only on the variation of temperature $\Delta T$. Since for an isothermal process $\Delta T=0$, then $\Delta U=0$. This means that the first law of thermodynamics becomes
$Q=W$
and since the work done is 370 J, then the amount of heat is also 370 J: $Q=370 J$.