Question

If equal mols of `"N"_2` and `"Ar"` are in the same container with a total pressure of `"10 atm"`, `(A)` what are their partial pressures? `(B)` what is the effusion speed of `"H"_2` if that of `"He"` is `"600 m/s"` at a certain temperature?

Answer 1

`A)` `"5 atm"` each.

`B)` The lighter gas `B` effuses at a speed `v_B` that is `sqrt(M_A/M_B)` times as fast as `v_A`, where `M_A` is the molar mass of `A`.

---

`A)` If the container is filled with an equal number of each particle, then we know that:

`n_(N_2) = n_(Ar)`

When this is the case, assuming ideal gases, their partial pressures should be identical.

Mathematically, this results in the following mol fractions:

`chi_(N_2) = (n_(N_2))/(n_(Ar) + n_(N_2))`

`chi_(Ar) = (n_(Ar))/(n_(Ar) + n_(N_2))`

But since `n_(N_2) = n_(Ar)`, we expect `chi_(N_2) = chi_(Ar) = 0.5`. Let's just say we had `"1 mol"` of each gas. Then:

`chi_(N_2) = 1/(1 + 1) = 0.5 = chi_(Ar)`

Assuming ideal gases, we can show that both gases have the same partial pressure, the pressure exerted by each gas in the mixture:

`color(blue)(P_(N_2)) = chi_(N_2)P_(t ot)`

`= chi_(Ar)P_(t ot) = color(blue)(P_(Ar))`

`= 0.5("10 atm") = color(blue)("5 atm")`

`B)` I'm not sure what part `B` has to do with part `A`. I will assume the question means to write `"Ar"` and `"N"_2` instead of `"He"` and `"H"_2`.

We are given, then, that argon effuses at `"600 m/s"`. Effusion can be derived from the expression for some sort of speed for each gas. The root-mean-square speed equation is fine:

`v_(rms) = sqrt((3RT)/M)`

where `R` and `T` are known from the ideal gas law, and `M` is the molar mass in `"kg/mol"`.

The rate of effusion, `z_(eff)` is proportional to the speed `v`, so the proportionality constants cancel out in a ratio:

`z_(eff,Ar)/(z_(eff,N_2)) = bb(v_(Ar)/(v_(N_2))) = bb(sqrt(M_(N_2)/(M_(Ar))))`

The relationship of `(z_(eff,i))/(z_(eff,j))` to `sqrt(M_j/M_i)` is known as Graham's law of effusion, but we will instead be using the relationship with `v_i/v_j`, the ratio of the speeds.

We are given that `v_(Ar) = "600 m/s"`, so:

`color(blue)(v_(N_2)) = v_(Ar) sqrt(M_(Ar)/M_(N_2))`

`= "600 m/s" cdot sqrt("0.039948 kg/mol"/"0.028014 kg/mol")`

`=` `color(blue)("716.5 m/s")`

This should make sense, that the lighter gas, `"N"_2`, effused faster.

If it really is supposed to be `"He"` and `"H"_2`, then you should expect:

`color(blue)(v_(H_2)) = v_(He) sqrt(M_(He)/M_(H_2))`

`= "600 m/s" cdot sqrt("0.0040026 kg/mol"/"0.0020158 kg/mol")`

`=` `color(blue)("845.5 m/s")`