What is the volume, in liters, occupied by 0.723 mol of a gas at 32.0°C if the pressure is 852 mm Hg?

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What is the volume, in liters, occupied by 0.723 mol of a gas at 32.0°C if the pressure is 852 mm Hg?

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Answer 1 · 2016-06-10T16:02:20

the volume will be $V = 16.2 L$ $V = 16.2L$

Explanation:

To solve for the volume we can use the Ideal Gas Law equation

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

$P =$ $P=$ Pressure in $a t m$ $atm$
$V =$ $V=$ Volume in $L$ $L$
$n =$ $n=$ moles
$R =$ $R=$ the Gas Law Constant $0.0821 \frac{a t m \cdot L}{m o l \cdot K}$ $0.0821 (atm*L)/(mol*K)$
$T =$ $T=$ Temperature in $K$ $K$

When using the Ideal Gas Law Constant we must convert the Pressure to $a t m$ $atm$
$P = 852 m m H g \cdot \frac{1 a t m}{760 m m H g} = 1.12 a t m$ $P=852mmHg*(1atm)/(760mmHg)= 1.12 atm$

$V = ?$ $V=?$

$n = 0.723 m o l$ $n=0.723mol$

$R = 0.0821 \frac{a t m \cdot L}{m o l \cdot K}$ $R=0.0821 (atm*L)/(mol*K)$

When using the Ideal Gas Law Constant we must convert the Temperature to $K$ $K$
$T = {32.0}^{o} C + 273 = 305 K$ $T=32.0^oC+273 = 305K$

$V = \frac{n R T}{P}$ $V=(nRT)/P$

$V = \frac{(0.723 m o l) (0.0821 \frac{a t m \cdot L}{m o l \cdot K}) (305 K)}{1.12 a t m}$ $V=((0.723cancel(mol))(0.0821 (cancel(atm)*L)/(cancel(mol)*cancel(K)))(305cancelK))/(1.12cancel(atm))$

$V = 16.2 L$ $V = 16.2L$

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What is the volume, in liters, occupied by 0.723 mol of a gas at 32.0°C if the pressure is 852 mm Hg?

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