$\\ Using\quad Charles'\quad Law\\ \\ \frac { { V }_{ 1 } }{ { T }_{ 1 } } =\quad \frac { { V }_{ 2 } }{ { T }_{ 2 } } \\ provided\quad the\quad pressure\quad of\quad the\quad gas\quad is\quad kept\quad constant.\\ Temperature\quad should\quad be\quad in\quad Kelvins:\quad \\ Kelvin\quad Temperature\quad =\quad Celsius\quad Temperature\quad +\quad 273.15\quad \\ Given\quad information:\\ { V }_{ 1 }=\quad 7L\quad \quad \quad \quad \quad \quad { V }_{ 2}\quad is\quad unknown\quad \quad {T}_{ 1 }=\quad 24+273.15=297.5K\quad \quad \quad { T }_{ 2 }=104+273.15=377,15K\\ \therefore \quad \frac { 7L }{ 297.5K } =\frac { { V }_{ 2 } }{ 377.5K } \\ \quad \quad \quad 2640.05\quad =\quad 297.5{ V }_{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad { V }_{2}=\frac { 2640.05 }{ 297.5 } \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad { V }_{ 2 }=8.9\quad L\quad to\quad the\quad nearest\quad tenth\quad of\quad a\quad liter.$