A container with a volume of #8 L# contains a gas with a temperature of #270^o C#. If the temperature of the gas changes to #360 ^o K# without any change in pressure, what must the container's new volume be?

Answer 1

#V_2=5.30L#

#V_1=8L# #T_1=270 ^o C=270+273=543 ^o K#
#T_2=360 ^o K# #V_2=?#
#V_1/T_1=V_2/T_2#
#8/543=V_2/360#
#V_2=(8*360)/543#
#V_2=5.30L#
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Answer 2

To find the new volume of the gas, we can use the combined gas law, which states:

[ \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2} ]

Given:

  • Initial volume ((V_1)) = 8 L
  • Initial temperature ((T_1)) = 270°C = 543 K
  • Final temperature ((T_2)) = 360 K (since temperature in Kelvin and Celsius scales have the same interval)
  • Pressure ((P_1) and (P_2)) remains constant

[ \frac{8 \times 543}{543} = \frac{8 \times V_2}{360} ]

[ V_2 = \frac{8 \times 360}{543} ]

[ V_2 ≈ 5.32 , 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.

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.

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.

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.

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