Question

If the volume of the container is `"1.00 L"`, what is the change in mols of gas required to change the pressure from `"1.00 atm"` to `"0.920 atm"` if the temperature was also dropped from `25^@ "C"` to `22^@ "C"`? (`R = "0.082057 L"cdot"atm/mol"cdot"K"`)

Answer 1

I get about `Deltan = n_2 - n_1 = "0.00289 mols"`.

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Since you are given:

• `V = "1.00 L"`, the volume
• `P_1 = "1.00 atm"`, the initial pressure
• `P_2 = "0.920 atm"`, the final pressure
• `T_1 = 25^@ "C"`, the initial temperature
• `T_2 = 22^@ "C"`, the final temperature
• `R = "0.082057 L"cdot"atm/mol"cdot"K"`, the universal gas constant

we know that the volume had to have remained constant (we only got one value for it).

Also, since we were given pressure, temperature, volume, and `R`, it's a good hint that we are using the ideal gas law:

`PV = nRT`

Since we have the initial and final values for pressure and temperature, we have two variations on the ideal gas law:

`P_1V = n_1RT_1`
`P_2V = n_2RT_2`

So finding "the mols" should mean finding the change in the mols, `n_2 - n_1 = Deltan`.

A reasonable way to do this is to solve for the initial `"mol"`s:

`n_1 = (P_1V)/(RT_1)`

`= (("1.00 atm")("1.00 L"))/(("0.082057 L"cdot"atm/mol"cdot"K")("25 + 273.15 K"))`

`=` `"0.0409 mols"`

So, `n_2` is gotten as follows:

`n_2 = (P_2V)/(RT_2)`

`= (("0.920 atm")("1.00 L"))/(("0.082057 L"cdot"atm/mol"cdot"K")("22 + 273.15 K"))`

`=` `"0.0380 mols"`

So, if you wanted the change in `"mol"`s, you would get:

`color(blue)(Deltan)`

`=` `"0.0409 - 0.0380 mols"`

`=` `color(blue)("0.00289 mols")`