Question

which container shown here has the smallest partial pressure of ar and why?

Explanation:

Step1: Recall Dalton's law of partial pressure

Partial pressure of a gas \(P_i = x_iP_{total}\), where \(x_i=\frac{n_i}{n_{total}}\) (mole fraction). For argon (\(Ar\)), if we assume the total pressure is the same for both containers (same conditions, volume, temperature, etc. implied by the problem - comparing relative amounts), the partial pressure of \(Ar\) depends on its mole fraction.

Step2: Count the number of \(Ar\) and total gas particles

In container \(a\):
Number of \(Ar\) particles (\(n_{Ar,a}\)) \(= 6\), total particles (\(n_{total,a}\)) \(=6 + 3=9\). Mole fraction \(x_{Ar,a}=\frac{6}{9}=\frac{2}{3}\).
In container \(c\):
Number of \(Ar\) particles (\(n_{Ar,c}\)) \(=3\), total particles (\(n_{total,c}\)) \(=3+7 = 10\). Mole fraction \(x_{Ar,c}=\frac{3}{10}\).
Since \(\frac{3}{10}<\frac{2}{3}\) (because \(3\times9 = 27\) and \(2\times10=20\), \(27<30\) so \(\frac{3}{10}=0.3\) and \(\frac{2}{3}\approx0.67\)).

Answer:

Container \(c\) has the smallest partial pressure of \(Ar\) because the mole - fraction of \(Ar\) (number of \(Ar\) particles divided by total number of particles) is smaller in container \(c\) compared to container \(a\). Using the formula \(P_{Ar}=x_{Ar}P_{total}\) (assuming same \(P_{total}\) for both containers as they are under comparable conditions - same volume and temperature implied), a smaller \(x_{Ar}\) (mole - fraction of \(Ar\)) leads to a smaller \(P_{Ar}\) (partial pressure of \(Ar\)).