Answer
The new temperature of the gas will be $~43.9~^{\circ}C$
Work Step by Step
We can find the number of moles of nitrogen:
$PV = nRT$
$n = \frac{PV}{RT}$
$n = \frac{(3.5)(1.013\times 10^5~Pa)(425\times 10^{-3}~m^3)}{(8.314~J/mol~K)(296.15~K)}$
$n = 61.2~mol$
We can find the change in temperature:
$Q = c~n~\Delta T$
$\Delta T = \frac{Q}{c~n}$
$\Delta T = \frac{26,600~J}{(20.8~J/mol~C^{\circ})(61.2~mol)}$
$\Delta T = 20.9~C^{\circ}$
The new temperature of the gas will be $23^{\circ}C+20.9~C^{\circ}$ which is $43.9~^{\circ}C$.
