Question

Question #359b2

Answer 1

`23^@"C"`

Explanation:

In order to find the temperature of the gas, you need to use the ideal gas law equation, which looks like this

`color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "`, where

`P` - the pressure of the gas
`V` - the volume it occupies
`n` - the number of moles of gas
`R` - the universal gas constant, usually given as `0.0821("atm" * "L")/("mol" * "K")`
`T` - the absolute temperature of the gas

The ideal gas law equation establishes a relationship between the pressure of the gas and the volume it occupies, on one hand, and the number of moles of gas and the absolute temperature of the gas, on the other.

It's important to keep in mind that you can only use the ideal gas law equation if the units given to you for pressure, volume, and temperature match those used in the expression of the universal gas constant.

In your case, all the units math those used for `R`, which means that you can rearrange the ideal gas law equation and solve for `T`, the absolute temperature of the gas

`PV = nRT implies T = (PV)/(nR)`

Plug in your values to get

`T = (1.0 color(red)(cancel(color(black)("atm"))) * 85color(red)(cancel(color(black)("L"))))/(3.5color(red)(cancel(color(black)("moles"))) * 0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * "K")) = "295.81 K"`

In order to express this temperature in degrees Celsius, you must use the conversion factor

`color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))`

In your case, you have

`t = "295.81 K" - 273.15 = 22.66^@"C"`

Rounded to two sig figs, the answer will be

`t = color(green)(|bar(ul(color(white)(a/a)23^@"C"color(white)(a/a)|)))`