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 = (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`.