Question

Question #d0747

Answer 1

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.