A container with a volume of #35 L# contains a gas with a temperature of #420^o K#. If the temperature of the gas changes to #300 ^o K# without any change in pressure, what must the container's new volume be?

We apply Charles' Law

Using the combined gas law, (PV/T = constant), where P is pressure, V is volume, and T is temperature:

(V_1 / T_1 = V_2 / T_2)

Given (V_1 = 35 , \text{L}), (T_1 = 420 , \text{K}), and (T_2 = 300 , \text{K}):

(35 , \text{L} / 420 , \text{K} = V_2 / 300 , \text{K})

Solve for (V_2):

(V_2 = (35 , \text{L} \times 300 , \text{K}) / 420 , \text{K} = 25 , \text{L})

Therefore, the container's new volume should be (25 , \text{L}).

To find the new volume of the container when the gas temperature changes from 420 K to 300 K without any change in pressure, you can use Charles's Law, which states that the volume of a given amount of gas is directly proportional to its absolute temperature if pressure and the amount of gas are constant.

Using Charles's Law, you can set up the following proportion:

[\frac{V_1}{T_1} = \frac{V_2}{T_2}]

Where: (V_1) = Initial volume of the gas (T_1) = Initial temperature of the gas (V_2) = New volume of the gas (which we need to find) (T_2) = New temperature of the gas

Given: (V_1 = 35) L (T_1 = 420) K (T_2 = 300) K

Plug in the given values into the equation:

[\frac{35}{420} = \frac{V_2}{300}]

Now, solve for (V_2):

[V_2 = \frac{35 \times 300}{420}]

[V_2 = \frac{10500}{420}]

[V_2 = 25]

So, the new volume of the container when the gas temperature changes to 300 K without any change in pressure must be 25 L.