How many moles of gas are contained in .89 L at 21.0 oC and .987 atm pressure?

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Answer 1 · 2018-03-03T04:32:34

$0.03609mol$

We shall assume that this is an Ideal Gas, and can thus use the Ideal Gas Law:

$PV=nRT$

where $P=$ Pressure (pascals)

where $V=$ Volume ( $m^3$ )

where $n=$ moles (mol)

where $T=$ Temperature (Kelvin)

and $R$ is the gas constant $8.31441 J K^-1 mol^-1$

Now, we convert the units:

$0.987atm=100007.775pa$

$0.89L=0.89dm^3=0.00089m^3$

$21^oC=294.15K$

Inputting the values into the equation:

$->100007.775 * 0.00089 = n * 8.31441 * 294.15$

$-> 89.0069=2445.6837n$

Solve for n:

$n= 89.0069/2445.6837$

$n=0.03639mol$

Thus, solved.