Question

What volume does 0.0250 mole `H_2` occupy at 0.821 atm pressure and 300 K?

Answer 1

The volume is `"0.750 dm"^3`

Explanation:

Hydrogen behaves like an ideal gas.

`P*V=n*R*T`, where

`P = "the pressure" = 0.821 color(red)(cancel(color(black)("atm")))*(1.01325*10^5 "Pa")/(1 color(red)(cancel(color(black)("atm"))))`

`P = "83 190" color(red)(cancel(color(black)("Pa"))) * "1N/m"^2/(1 color(red)(cancel(color(black)("Pa")))) = "83 190 N/m"^2`

`T= "the temperature"`

`n = "number of moles"`

`R= "ideal gas constant" = 8.314 color(red)(cancel(color(black)("J")))//("K·mol") * "1 N·m"/(1 color(red)(cancel(color(black)("J")))) = "8.314 N·m/(K·mol)"`

`V = "the volume"`

`V=(n*R*T)/P`

`V=((0.250 color(red)(cancel(color(black)("mol"))) * 8.314 color(red)(cancel(color(black)("N")))*"m")//(color(red)(cancel(color(black)("K·mol")))) · 300 color(red)(cancel(color(black)("K"))))/("83 190" color(red)(cancel(color(black)("N")))//"m"^2)`

`V = 7.50*10^-4 " m"^3 = "0.750 dm"^3`

Answer 2

The volume will be `"0.7 L"`.

Explanation:

Use the ideal gas law.

`PV=nRT`, where `n` is moles, and `R` is the gas constant.

Given/Known
`P="0.821 atm"`
`n="0.0250 mol"`
`R="0.082057338 L atm K"^(-1) "mol"^(-1)`
`T="300 K"`

Unknown

`V`

Equation

`PV=nRT`

Solution
Rearrange the equation to isolate `V` and solve.

`V=(nRT)/P`

`V=((0.0250cancel"mol"xx0.08205733" L" cancel"atm" cancel("K"^(-1)) cancel("mol"^(-1)) xx 300cancel"K"))/(0.821cancel"atm")="0.7 L"` (rounded to one significant figure due to 300 K)