Question

nitrogen gas was collected by water displacement. what was the pressure of N2 collected if the temperature was at 22 degrees calcius

Answer 1

Therefore, the pressure of N2 collected by water displacement at a temperature of 22 degrees Celsius (295.15 Kelvin) would be approximately 101.3 kPa, assuming the volume remains constant during the collection process.

The pressure of a gas collected by water displacement can be determined using the combined gas law, which relates the initial pressure, volume, and temperature of a gas to its final pressure, volume, and temperature.

The combined gas law equation is:

`\[ P_1 * V_1 / T_1 = P_2 * V_2 / T_2 \]`

Where:

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`\( P_1 \)` is the initial pressure of the gas.

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`\( V_1 \)` is the initial volume of the gas.

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`\( T_1 \)` is the initial temperature of the gas in Kelvin.

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`\( P_2 \)` is the final pressure of the gas (in this case, the pressure of N2 collected).

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`\( V_2 \)` is the final volume of the gas.

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`\( T_2 \)` is the final temperature of the gas in Kelvin.

Given that the temperature is 22 degrees Celsius, let's convert it to Kelvin:

`\[ T(K) = T(C) + 273.15 \]`

`\[ T(K) = 22 + 273.15 = 295.15 \, \text{K} \]`

Assuming the volume remains constant, as the gas is collected by water displacement, the initial pressure of the gas can be equated to the pressure of the atmosphere, which is typically around 1 atmosphere or 101.3 kPa.

So, rearranging the combined gas law equation to solve for
`\( P_2 \):`

`\[ P_2 = P_1 * (V_2 * T_1)/(V_1 * T_2) \]`

Substituting the values:

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`\( P_1 = 101.3 \, \text{kPa} \)` (standard atmospheric pressure)

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`\( V_2 = V_1 \)` (assuming the volume remains constant)

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`\( T_1 = 295.15 \, \text{K} \)`

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`\( T_2 = 295.15 \, \text{K} \)` (same temperature as initial)

`\[ P_2 = 101.3 * (V_1 * 295.15)/(V_1 * 295.15) \]`

`\[ P_2 = 101.3 \, \text{kPa} \]`