3. A pipe has an effusion leak from a small hole. The gas temperature is 30 °C. The molar leak rate is $\dot{n}$ (mols/h). The gas temperature changes to 55 °C. By what factor does the leak rate change? Answer as a ratio of the new leak rate to the old leak rate.

(Note: Provide the answer in the format of a ratio.)

When the gas temperaturey proportional to the square root of its temperature. In this case, we have the same gas leaking through the hole, so the molar mass remains constant. Therefore, the leak rate is directly proportional to the square root of the temperature.

Initially, the gas is at 30°C, which is equivalent to 303.15 Kelvin. At this temperature, let's call the initial leak rate ǹ1. When the temperature increases to 55°C (328.15 Kelvin), the new leak rate, ǹ2, can be calculated as follows:

([tex]ǹ2 / ǹ1) = √(T2 / T1)[/tex]

[tex](ǹ2 / ǹ1) = √(328.15 / 303.15)[/tex]

[tex](ǹ2 / ǹ1) ≈ 1.183[/tex]

So, the leak rate changes by a factor of approximately 1.83 (1.183 rounded to two decimal places), with the new leak rate being 1.83 times the old leak rate.

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