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A container with a volume of #12 L# contains a gas with a temperature of #230^o C#. If the temperature of the gas changes to #420 ^o K# without any change in pressure, what must the container's new volume be?

Answer 1

Charles Aeschliman

I get approximately #10 \ "L"#.

We first convert #230^@"C"# into #"K"#, and we have:

#230^@"C"=(273.15+230) \ "K"# (since #"K"=""^@"C"+273.15#)

#=503.15 \ "K"#

If there is no change in pressure and the number of moles, we may use Charles's law, which states that

#VpropT#

#V_1/T_1=V_2/T_2#

We need to solve for the new volume, so we rearrange the equation in terms of #V_2#, and we get:

#V_2=V_1/T_1*T_2#

Plugging in the given values, we find that

#V_2=(12 \ "L")/(503.15color(red)cancelcolor(black)"K")*420color(red)cancelcolor(black)"K"#

#~~10 \ "L"# to the nearest whole number.

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Answer 2

Julia Aguilar

To find the new volume of the gas, use the combined gas law equation: ( V_2 = \frac{{V_1 \times T_2}}{{T_1}} ). Substitute the given values into the equation: ( V_2 = \frac{{12 \times 420}}{{230 + 273.15}} ). Calculate ( V_2 ) to find the new volume of the gas.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.