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

Nov 20, 2015

The volume is ${\text{0.750 dm}}^{3}$

#### Explanation:

Hydrogen behaves like an ideal gas.

$P \cdot V = n \cdot R \cdot T$, where

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

$P = {\text{83 190" color(red)(cancel(color(black)("Pa"))) * "1N/m"^2/(1 color(red)(cancel(color(black)("Pa")))) = "83 190 N/m}}^{2}$

$T = \text{the temperature}$

$n = \text{number of moles}$

$R = \text{ideal gas constant" = 8.314 color(red)(cancel(color(black)("J")))//("K·mol") * "1 N·m"/(1 color(red)(cancel(color(black)("J")))) = "8.314 N·m/(K·mol)}$

$V = \text{the volume}$

$V = \frac{n \cdot R \cdot T}{P}$

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)

$V = 7.50 \cdot {10}^{-} 4 {\text{ m"^3 = "0.750 dm}}^{3}$

Nov 20, 2015

The volume will be $\text{0.7 L}$.

#### Explanation:

Use the ideal gas law.

$P V = n R T$, where $n$ is moles, and $R$ is the gas constant.

Given/Known
$P = \text{0.821 atm}$
$n = \text{0.0250 mol}$
$R = {\text{0.082057338 L atm K"^(-1) "mol}}^{- 1}$
$T = \text{300 K}$

Unknown

$V$

Equation

$P V = n R T$

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

$V = \frac{n R T}{P}$

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)