Question #d0747

1 Answer
May 16, 2017

Here's what I got.

Explanation:

The idea here is that the volume of a gas varies directly with its temperature when the number of moles of gas and the pressure are kept constant #-># this is known as Charles' Law.

You can describe this as

#color(blue)(ul(color(black)(V_1/T_1 = V_2/T_2)))#

Here

  • #V_1# and #T_1# represent the volume and the temperature of the gas at an initial state
  • #V_2# and #T_2# represent the volume and the temperature of the gas at a final state

In your case, the volume of the gas is increasing

#"0.43 mL " ->" 1 mL"#

which means that the temperature must have increased.

Rearrange the equation to solve for #T_2#, the temperature at which the gas occupies #"1 mL"#

#V_1/T_1 = V_2/T_2 implies T_2 = V_2/V_1 * T_1#

Plug in your values to find

#T_2 = (1 color(red)(cancel(color(black)("mL"))))/(0.43color(red)(cancel(color(black)("mL")))) * "299 K" = "695.3 K"#

To convert this to degrees Celsius, use the fact that

#color(blue)(ul(color(black)(t[""^@"C"] = T["K"]- 273.15)))#

You will end up with

#t[""^@"C"] = "695.3 K" - 273.15 = color(darkgreen)(ul(color(black)(420^@"C")))#

I'll leave the answer rounded to two sig figs, but keep in mind that only have one significant figure for the second volume of the gas.