## College Physics (4th Edition)

There are $~4.83\times 10^{21}~$ gas molecules in the chamber.
We can find the number of moles $n$: $Q = c_v~n~\Delta T$ $n = \frac{Q}{c_v~\Delta T}$ $n = \frac{Q}{1.5~R~\Delta T}$ $n = \frac{10.0~J}{(1.5)(8.314~J/mol~K)(100~K)}$ $n = 0.00802~mol$ We can find the number of molecules: $N = (0.00802~mol)(6.022\times 10^{23}) = 4.83\times 10^{21}$ There are $~4.83\times 10^{21}~$ gas molecules in the chamber.