Question

A gas sample occupies 9.96 L at 3.0C. What is the pressure given that there are 1.26 mol of gas in the sample?

Answer 1

`2.87 " atm"`

Explanation:

Using the ideal gas law: `PV=nRT`

`P= (nRT)/V`

The temperature has to be in Kelvin:
`3^@C+273= 276 K`

`1.26 cancel"mol"xx(0.0821 "Latm")/(cancel"mol"cancel"K")xx(276cancel"K")/(9.96cancel"L")= 2.87 "atm"`

Answer 2

The pressure will be `"290. kPa"`, `"0.290 Pa"`, or `"2.87 atm"`, depending on the gas constant used.

Explanation:

Use the ideal gas law :

`PV=nRT`,

where:

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

The Celsius temperature will need to be converted to Kelvins by adding `273.15`.

Known

`V="9.96 L"`

`n="1.26 mol"`

`R="8.31447 L kPa K"^(-1) "mol"^(-1)`

`T="3.0"^@"C + 273.15"="276.2 K"`

Unknown

`P`

Solution

Rearrange the equation to isolate `P`. Plug in the known values and solve.

`P=(nRT)/V`

`P=(1.26color(red)cancel(color(black)("mol"))xx8.31447color(red)cancel(color(black)(" L")) "kPa" color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx276.2color(red)cancel(color(black)("K")))/(9.96color(red)cancel(color(black)("L")))="290. kPa"` (rounded to three significant figures)

`"290. kPa"` `=` `"0.290 Pa"`

If your instructor requires the pressure to be in atm, use `"0.08206 L atm K"^(-1) "mol"^(-1)` instead of `"8.31447 L kPa K"^(-1) "mol"^(-1)` for the gas constant `R`. The answer will be `"2.87 atm"`.