# A sample of hydrogen gas occupies a volume of 33.4 L at STP. How many moles of hydrogen gas is this?

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1
Feb 9, 2018

I got $\text{1.47 mols}$....

If we ASSUME (!!!!) that hydrogen gas is an ideal gas, then we ASSUME that the ideal gas law works:

$P V = n R T$

• $P$ is pressure in $\text{atm}$ or $\text{bar}$...
• $V$ is volume in $\text{L}$.
• $R = \text{0.083145 L"cdot"bar/mol"cdot"K}$ $=$ $\text{0.082057 L"cdot"atm/mol"cdot"K}$ is the universal gas constant.
• $T$ is the temperature in $\text{K}$.
• $n$ is obviously the mols of IDEAL gas.

And so,

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

STP is defined since 1982 as ${0}^{\circ} \text{C}$ and $\text{1 bar}$. STP is defined before 1982 as ${0}^{\circ} \text{C}$ and $\text{1 atm}$.

Before 1982,

$n = \left(\text{1 atm"cdot"33.4 L")/("0.082057 L"cdot"atm/mol"cdot"K" cdot "273.15 K}\right)$

$=$ $\text{1.49 mols}$

But since we're not old-timers who are stuck in the past, we look at AFTER 1982 to obtain:

$n = \left(\text{1 bar" cdot "33.4 L")/("0.083145 L"cdot"bar/mol"cdot"K" cdot "273.15 K}\right)$

$=$ $\underline{\textcolor{b l u e}{\text{1.47 mols}}}$

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#### Explanation

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#### Explanation:

I want someone to double check my answer

1
Feb 9, 2018

At STP, all gases have a molar volume of 22.4 L/mol
$\left(33.4 L\right) \left(\frac{1 m o l}{22.4 L}\right) = 1.49 m o l$

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