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?
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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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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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