8.00 L of a gas is collected at 60.0°C. What will be its volume upon cooling to 30.0°C?

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8.00 L of a gas is collected at 60.0°C. What will be its volume upon cooling to 30.0°C?

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Answer 1 · 2014-12-07T10:20:03

The gas' new volume will be $V_{2} = 7.28$ $V_2 = 7.28$ L.

This problem is a simple application of Charles' law

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1 = V_2/T_2$ ,

which states that the volume of a gas is directly proportional to its temperature(measured in $K$ $K$ - very important).

So, we have an initial volume equal to $V_{1} = 8.00 L$ $V_1 = 8.00L$ , an initial temperature of $T_{1} = (273.15 + 60) = 333.15 K$ $T_1 = (273.15+60) = 333.15K$ , and a final temperature of $T_{2} = (273.15 + 30) = 303.15 K$ $T_2 = (273.15 + 30) = 303.15K$ , which gets us

$V_{2} = \frac{T_{2}}{T_{1}} \cdot V_{1} = \frac{303.15}{333.15} \cdot 8.00 = 7.28 L$ $V_2 = T_2/T_1 * V_1 = 303.15/333.15 * 8.00 = 7.28L$ -> a decrease in temperature is correlated with a decrease in volume, just as expected.

Note, however, what would have happened with degrees Celsius instead of K:

$V_{2} = \frac{30}{60} \cdot 8.00 = 4.00 L$ $V_2 = 30/60 * 8.00 = 4.00 L$ , a result significantly different...

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8.00 L of a gas is collected at 60.0°C. What will be its volume upon cooling to 30.0°C?

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