Question #9e2f2

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Question #9e2f2

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Answer 1 · 2015-03-12T18:25:38

The temperature will be $570 K$ $"570 K"$ .

Since no mention of pressure and number of moles was made, you can safely assume that they're kept constant, which means that you can use the equation for Charles' Law.

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

$V_{1}$ $V_1$ , $T_{1}$ $T_1$ - the initial volume and the initial temperature;
$V_{2}$ $V_2$ , $T_{2}$ $T_2$ - the final volume and temperature;

So, you know the your balloon occupied a volume of 3.4 L at 283 K. Even before using the equation, you can predict what's going to happen to the temperature.

Because pressure and number of moles are kept constant, an increase in the volume of the ballonn could only come about with an increase in the temperature of the gas.

Since a higher temperature means a bigger average kinetic energy for the gas particles, the force exercited on the walls of the balloon increases $\to$ $->$ the balloon expands.

![http://wps.prenhall.com/wps/media/objects/476/488316/ch11.html]( useruploads.socratic.org )

So, numerically, this means that

$T_{2} = \frac{V_{2}}{V_{1}} \cdot T_{1} = \frac{6.9 L}{3.4 L} \cdot 283 K = 574.3 K$ $T_2 = V_2/V_1 * T_1 = "6.9 L"/"3.4 L" * "283 K" = "574.3 K"$

Rounded to two sig figs, the number of sig figs in 3.4 and 6.9 L, the answer will be

$T_{2} = 570 K$ $T_2 = "570 K"$

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Question #9e2f2

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