A gas at 69°C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.86 L?

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A gas at 69°C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.86 L?

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Answer 1 · 2015-03-06T18:57:19

The volume will increase to 1.86 L at 680 °C.

This is a problem involving Charles' Law.

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

In your problem,

$V_{1} = 0. 67 L$ $V_1 = "0. 67 L"$ ; $V_{2} = 1.86 L$ $V_2 = 1. 86 L"$
$T_{1} = (69 + 273.15) K = 342.15 K$ $T_1 = "(69 + 273.15) K" = "342.15 K"$ ; $T_{2} = ?$ $T_2 = "?"$

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

$T_{2} = T_{1} \times \frac{V_{2}}{V_{1}} = 342.15 K \times \frac{1.86 L}{0.67 L} = 950 K$ $T_2 = T_1 × V_2/V_1 = "342.15 K" × "1.86 L"/"0.67 L" = "950 K"$ (2 significant figures)

$T_{2} = (950 - 273.15) °C = 680 °C$ $T_2 = "(950 – 273.15) °C" = "680 °C"$

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A gas at 69°C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.86 L?

1 Answer

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