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

`14{ 2 }^( 0 )C`

Step-by-step 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`