Question

At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres?

At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95 atmospheres?

Answer 1

`T=447 K~~174^@C`.

Explanation:

This is what we're given:

• `P` (pressure), which is `"1.95 atm"`.
• `V` (volume), which is `"12.30 L"`.
• `n` (number of moles), which is `0.654` moles of neon gas.

We have to find `T`, or temperature. To do this, we'll need to use the Ideal Gas Law, which is:

`PV = nRT`

Rearranging this equation to get temperature on one side, we get:

`T = (PV)/(nR)`

Pressure is in `atm` and volume is in `L`. This tells us that we'll need to use the value of `"0.08206 L atm/K mol"` for `R`, the ideal gas constant.

Plugging in all of the values, we can solve for temperature:

`T = (pV)/(nR)`

`T = (1.95color(red)cancelcolor(black)"atm" xx 12.30 color(red)cancelcolor(black)"L")/(0.654color(red)cancelcolor(black)"mol" xx 0.08206color(red)cancelcolor(black)Lcolor(red)cancelcolor(black)"atm"K^-1 color(red)cancelcolor(black)"mol")`

`T = 447 K`

We can then convert to celsius, which equals to around `174^@C`.

Answer 2

The temperature of neon gas under the conditions given is `"477 K"`.

Explanation:

Use the equation for the ideal gas law:

`PV=nRT`,

where:

`P` is pressure, `V` is volume, `n` is mole, `R` is the gas constant, `T` is temperature in Kelvins.

Organize data:

Known

`P="1.95 atm"`

`V="12.30 L"`

`n="0.654 mol"`

`R="0.0820575 L atm K"^(-1) "mol"^(-1)`

Unknown

`T`

Solution

Rearrange the ideal gas equation to isolate `T`. Plug in known values and solve.

`T=(PV)/(nR)`

`T=(1.95color(red)cancel(color(black)("atm"))xx12.30color(red)cancel(color(black)("L")))/(0.654color(red)cancel(color(black)("mol"))xx0.0820575 color(red)cancel(color(black)("L")) color(red)cancel(color(black)("atm")) "K"^(-1) color(red)cancel(color(black)("mol"^(-1))))="447 K"` (rounded to three significant figures)