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A container with a volume of #25 L# contains a gas with a temperature of #150^o K#. If the temperature of the gas changes to #320 ^o K# without any change in pressure, what must the container's new volume be?

Answer 1

Layla Ahmed

#160/3L#

From Ideal Gas Equation; #PV=nRT#

Pressure, number of moles and Gas constant are all fixed.

#V=nRT/P# or #V/T=nR/P#

As long as pressure and the amount of gas remains unchanged, the ratio of #V# and #T# is always the same.

#V_1/T_1=V_2/T_2# #(25L)/(150K)=(V_2)/(320K)# #V_2=160/3L#

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Answer 2

Carson Alexander

Using the combined gas law, $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ , and given that pressure remains constant:

$\frac{V_2}{V_1} = \frac{T_2}{T_1}$

Plugging in the values:

$\frac{V_2}{25\, \text{L}} = \frac{320\, \text{K}}{150\, \text{K}}$

$V_2 ≈ \frac{320\, \text{K}}{150\, \text{K}} \times 25\, \text{L}$

$V_2 ≈ 53.33\, \text{L}$

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Answer from HIX Tutor

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