What volume (L) will 0.20 mol HI occupy at 300 K and 100.0 kPa? R = 8.314 kPa L / (K mol) = 0.08205 atm L / (mol K)?

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What volume (L) will 0.20 mol HI occupy at 300 K and 100.0 kPa? R = 8.314 kPa L / (K mol) = 0.08205 atm L / (mol K)?

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Answer 1 · 2014-12-15T18:35:38

The answer is $5 L$ $5L$ .

You can use the ideal gas law, $P V = n R T$ $PV = nRT$ , to determine the volume, given the fact that all other variables are known.

First, let's use the equation that has $R = 0.08205 \frac{a t m \cdot L}{m o l \cdot K}$ $R = 0.08205 (atm * L)/(mol * K)$ . This means that we need to convert the $100.0 k P a$ $100.0kPa$ pressure to $a t m$ $atm$

$100.0 k P a \cdot \frac{1 a t m}{101.325 k P a} = 0.9869 a t m$ $100.0 kPa * (1 atm)/(101.325 kPa) = 0.9869atm$

So, $V = \frac{n R T}{P} = \frac{0.2 \cdot 0.08205 \cdot 300}{0.9869} = 5 L$ $V = (nRT)/P = (0.2 * 0.08205 * 300)/(0.9869) = 5L$ (rounded to 1 sig fig);

As practice and as a way to double-check our result, use the other value of $R$ $R$ as well; this time, the pressure stays in $k P a$ $kPa$ and the volume must come out to be $5 L$ $5L$

$V = \frac{n R T}{P} = \frac{0.2 \cdot 8.314 \cdot 300}{100.0} = 5 L$ $V = (nRT)/P = (0.2 * 8.314 * 300)/(100.0) = 5L$

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What volume (L) will 0.20 mol HI occupy at 300 K and 100.0 kPa? R = 8.314 kPa L / (K mol) = 0.08205 atm L / (mol K)?

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