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 #390 ^o K# without any change in pressure, what must the container's new volume be?

Answer 1

The volume is #=5.75L#

Charles' Law is employed.

#V_1/T_1=V_2/T_2#
#V_1=8L#
#T_1=270ºC+273=543K#
#T_2=390K#
#V_2=T_2/T_1*V_1#
#=390/543*8=5.75L#
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Answer 2

To find the new volume, use the equation:

V2 = V1 * (T2 / T1)

Where: V1 = initial volume (8 L) T1 = initial temperature in Kelvin (270°C + 273.15 = 543.15 K) T2 = final temperature in Kelvin (390 K)

Substitute the values into the equation:

V2 = 8 L * (390 K / 543.15 K) V2 ≈ 5.75 L

Therefore, the container's new volume must be approximately 5.75 liters.

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Answer 3

Using the ideal gas law equation (PV = nRT), where (P) is the pressure, (V) is the volume, (n) is the number of moles of gas, (R) is the gas constant, and (T) is the temperature in Kelvin, we can find the new volume of the gas.

Given that the pressure remains constant, we can rewrite the equation as (V_1 / T_1 = V_2 / T_2), where (V_1) is the initial volume, (T_1) is the initial temperature, (V_2) is the final volume, and (T_2) is the final temperature.

Plugging in the given values, we have: (V_1 / T_1 = V_2 / T_2), (V_1 / (270 + 273) = V_2 / 390), (V_1 / 543 = V_2 / 390).

Rearranging the equation to solve for (V_2): (V_2 = (V_1 / 543) * 390), (V_2 = (8 / 543) * 390), (V_2 ≈ 5.73 , L).

So, the container's new volume must be approximately 5.73 liters.

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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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