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A container with a volume of #5 L# contains a gas with a temperature of #320^o K#. If the temperature of the gas changes to #240^o K# without any change in pressure, what must the container's new volume be?

Answer 1

Layla Ahmed

#V_2=3,75 L#

#"first volume: " V_1=5L# #"first temperature :"T_1=320^o K#

#"last temperature :"T_2=240^o K# #"last volume :"V_2=?#

#V_1/T_1=V_2/T_2#

#V_2=(V_1*T_2)/T_1#

#V_2=(5*240)/320#

#V_2=3,75 L#

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Answer 2

Lily Adams

Use the combined gas law: [ P_1V_1/T_1 = P_2V_2/T_2 ] [ V_2 = V_1 \times \frac{T_2}{T_1} ] [ V_2 = 5 , \text{L} \times \frac{240}{320} ] [ V_2 = 3.75 , \text{L} ]

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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.