Question

An ideal gas initially held at a pressure of 1.1 bar and a temperature of 12°C has a volume of 4.0 L. If the gas is allowed to expand to a volume of 8.0 L while its pressure drops to 0.8 bar, what must be the final temperature of the gas to the nearest degree Celsius? Hint: remember to use the appropriate temperature scale!

Answer 1

Answer:

$14{ 2 }^{ 0 }C$

Explanation:

$\\ Using\quad the\quad general\quad gas\quad equation\\ \\ \frac { { { P }_{ 1 }V }_{ 1 } }{ { T }_{ 1 } } =\quad \frac { { { P }_{ 2 }V }_{ 2 } }{ { T }_{ 2 } } \\ Temperature\quad should\quad be\quad in\quad Kelvins:\quad \\ Kelvin\quad Temperature\quad =\quad Celsius\quad Temperature\quad +\quad 273.15\quad \\ Given\quad information:\\ { V }_{ 1 }=\quad 4L\quad { V }_{ 2\quad }=8L\quad { P }_{ 1 }=1.1bar\quad { P }_{ 2 }=0.8bar\quad \quad { T }_{ 1 }=\quad 12+273.15=285.15K\quad \quad \quad { T }_{ 2 }=?\quad \\ \therefore \quad \frac { 1.1(4) }{ 285.15 } =\frac { 0.8(8) }{ { T }_{ 2 } } \\ \quad \quad 1824.96\quad =\quad 4.4{ T }_{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad { T }_{ 2 }=\frac { 1824.96 }{ 4.4 } \quad \\ \quad \quad \quad \quad \quad \quad \quad { T }_{ 2 }=414.76K\\ {T}_{ 2}=(414.76-273.15{ ) }^{ 0 }C\\ \quad \quad \quad \quad \quad \quad \quad { T }_{2}=14{ 2 }^{ 0 }C$