A man heats a balloon in the oven. If the balloon initially has a volume of 4 liters and a temperature of 20°C, what will the volume of the balloon be after he heats it to a temperature of 250°C?

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A man heats a balloon in the oven. If the balloon initially has a volume of 4 liters and a temperature of 20°C, what will the volume of the balloon be after he heats it to a temperature of 250°C?

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Answer 1 · 2016-06-25T13:41:20

We use old Charles' Law. to get approximately $7 L$ $7 "L"$ .

Explanation:

Since, for a given quantity of gas, $V \propto T$ $VpropT$ if $P$ $P$ is constant, $V = k T$ $V=kT$ .

Solving for $k$ $k$ , $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1=V_2/T_2$ , and $V_{2} = \frac{V_{1} \times T_{2}}{T_{1}}$ $V_2=(V_1xxT_2)/T_1$ ; $T$ $T$ is reported in $degrees Kelvin$ $"degrees Kelvin"$ , $V$ $V$ may be in whatever units you like, $pints, sydharbs, gills, bushels etc.$ $"pints, sydharbs, gills, bushels etc."$ . Of course, we stick with sensible units, i.e. $L, litres$ $L," litres"$ .

Thus $V_{2}$ $V_2$ $=$ $=$ $\frac{4 L \times (250 + 273) K}{(20 + 273) K}$ $(4 "L"xx(250+273)K)/((20+273)K)$ $≅$ $~=$ $7 L$ $7 "L"$

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A man heats a balloon in the oven. If the balloon initially has a volume of 4 liters and a temperature of 20°C, what will the volume of the balloon be after he heats it to a temperature of 250°C?

1 Answer

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