What will be the volume of $8 g$ $8g$ of $H_{2}$ $H_2$ gas at the pressure of $4 a t m$ $4atm$ and temperature of $300 K$ $300K$ ?

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What will be the volume of $8 g$ $8g$ of $H_{2}$ $H_2$ gas at the pressure of $4 a t m$ $4atm$ and temperature of $300 K$ $300K$ ?

What will be the volume of $8 g$ $8g$ of $H_{2}$ $H_2$ gas at the pressure of $4 a t m$ $4atm$ and temperature of $300 K$ $300K$ .

$N O T E :$ $NOTE:$
take
$(H = 1)$ $(H=1)$
$(R = 0.082 L i t - a t m - m o l^{- 1} K^{- 1})$ $(R=0.082Lit-atm-mol^-1K^-1)$

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Answer 1 · 2018-01-11T13:01:35

$24.42 = 24420 c m^{3}$ $24.42=24420cm^3$

Explanation:

We are going to use the ideal gas law, which states that

$P V = n R T$ $PV=nRT$

$P$ $P$ is pressure in pascals $(P a)$ $(Pa)$

$V$ $V$ is volume in $m^{3}$ $m^3$

$n$ $n$ is the number of moles of the substance

$R$ $R$ is the gas constant

$T$ $T$ is the temperature in Kelvin $(K)$ $(K)$

Since we are using atmospheric pressure as units and the gas constant as $0.082 L i t - a t m - m o l^{- 1} K^{- 1}$ $0.082 Lit-atm-mol^-1K^-1$ , then our volume will be in liters $(l)$ $(l)$ .
So, plugging in $P = 4 a t m$ $P=4atm$ , $n = \frac{8g}{2.016g/mol} \approx 3.97 mol$ $n="8g"/"2.016g/mol"~~3.97 "mol"$ , $R = 0.082$ $R=0.082$ , and $T = 300 K$ $T=300K$ , we get

$V = \frac{n R T}{P} = \frac{3.97 \cdot 0.082 \cdot 300}{4} \approx 24.42 l = 24420 c m^{3}$ $V=(nRT)/P=(3.97*0.082*300)/4~~24.42l=24420cm^3$ of hydrogen gas $(H_{2})$ $(H_2)$ .

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What will be the volume of $8 g$ $8g$ of $H_{2}$ $H_2$ gas at the pressure of $4 a t m$ $4atm$ and temperature of $300 K$ $300K$ ?